MECHANICAL
ENGINEERING PRINCIPLES
John Bird & Carl Ross
THIRD EDITION
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Mechanical Engineering Principles
Third Edition
Why are competent engineers so vital?
Engineering is among the most important of all professions. It is the authors’ opinions that engineers save more
lives than medical doctors (physicians). For example, poor water, or the lack of it, is the second largest cause
of human death in the world, and if engineers are given the ‘tools’, they can solve this problem. The largest
cause of human death is caused by the malarial mosquito, and even death due to malaria can be decreased by
engineers - by providing helicopters for spraying areas infected by the mosquito and making and designing
medical syringes and pills to protect people against catching all sorts of diseases. Most medicines are produced
by engineers! How does the engineer put 1 mg of ‘medicine’ precisely and individually into millions of pills, at
an affordable price?
Moreover, one of the biggest contributions by humankind was the design of the agricultural tractor, which
was designed and built by engineers to increase food production many-fold, for a human population which
more-or-less quadruples every century! It is also interesting to note that the richest countries in the world are
very heavily industrialized. Engineers create wealth! Most other professions don’t!
Even in blue sky projects, engineers play a major role. For example, most rocket scientists are chartered
engineers or their equivalents and Americans call their chartered engineers (and their equivalents), scientists.
Astronomers are space scientists and not rocket scientists; they could not design a rocket to conquer outer
space. Even modern theoretical physicists are mainly interested in astronomy and cosmology and also nuclear
science. In general a theoretical physicist cannot, without special training, design a submarine structure to
dive to the bottom of the Mariana Trench, which is 11.52 km or 7.16 miles deep, or design a very long bridge, a
tall city skyscraper or a rocket to conquer outer space. It may be shown that the load on a submarine pressure
hull of diameter 10 m and length 100 m is equivalent to carrying the total weight of about 7 million London
double-decker buses!
This book presents a solid foundation for the reader in mechanical engineering principles, on which s/he
can safely build tall buildings and long bridges that may last for a thousand years or more. It is the authors’
experience that it is most unwise to attempt to build such structures on shaky foundations; they may come
tumbling down - with disastrous consequences.
John Bird is the former Head of Applied Electronics in the Faculty of Technology at Highbury College, Portsmouth,
U.K. More recently, he has combined freelance lecturing at the University of Portsmouth, with Examiner
responsibilities for Advanced Mathematics with City and Guilds, and examining for the International Baccalaureate
Organisation. He is the author of over 125 textbooks on engineering and mathematical subjects with worldwide
sales of one million copies. He is currently a Senior Training Provider at the Defence School of Marine Engineering
in the Defence College of Technical Training at H.M.S. Sultan, Gosport, Hampshire, U.K.
Carl Ross gained his first degree in Naval Architecture, from King’s College, Durham University; his PhD in
Structural Engineering from the Victoria University of Manchester; and was awarded his DSc in Ocean Engineering
from the CNAA, London. His research in the field of engineering led to advances in the design of submarine pressure
hulls. His publications and guest lectures to date exceed some 290 papers and books, etc., and he is Professor of
Structural Dynamics at the University of Portsmouth, UK.
See Carl Ross’s website below , which has an enormous content on science , technology and education .
http://tiny.cc/6kvqhx
Some quotes from Albert Einstein (14 March 1879-18 April 1955)
‘Scientists investigate that which already is; Engineers create that which has never been 9
‘Imagination is more important than knowledge. For knowledge is limited to all we now know and understand ,
while imagination embraces the entire world, and all there ever will be to know and understand 9
‘Everybody is a genius. But if you judge a fish by its ability to climb a tree, it will live its whole life believing
that it is stupid 9
‘To stimulate creativity, one must develop the childlike inclination for play 9
Mechanical Engineering Principles
Third Edition
John Bird BSc(Hons), CEng, CMath, CSci, FIMA, FIET, FCollT
Carl Ross BSc(Hons), PhD, DSc, CEng, FRINA, MSNAME
13 Routledge
Taylor &. Francis Group
LONDON AND NEW YORK
Third edition published 2015
by Routledge
2 Park Square, Milton Park, Abingdon, Oxon 0X14 4RN
and by Routledge
711 Third Avenue, New York, NY 10017
Routledge is an imprint of the Taylor & Francis Group, an informa business
©2015 John O. Bird and Carl T. F. Ross
The right of John O. Bird and Carl T. F. Ross to be identified as authors of this work has been asserted by them in
accordance with sections 77 and 78 of the Copyright, Designs and Patents Act 1988.
All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic,
mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any
information storage or retrieval system, without permission in writing from the publishers.
This publication presents material of a broad scope and applicability. Despite stringent efforts by all concerned in the
publishing process, some typographical or editorial errors may occur, and readers are encouraged to bring these to our
attention where they represent errors of substance. The publisher and author disclaim any liability, in whole or in part,
arising from information contained in this publication. The reader is urged to consult with an appropriate licensed
professional prior to taking any action or making any interpretation that is within the realm of a licensed professional
practice.
First edition published by Elsevier in 2002
Second edition published by Routledge in 2012
Trademark notice : Product or corporate names may be trademarks or registered trademarks, and are used only for
identification and explanation without intent to infringe.
British Library Cataloguing in Publication Data
A catalogue record for this book is available from the British Library
Library of Congress Cataloguing-in-Publication Data
Bird, J. O.
Mechanical engineering principles / John Bird and Carl Ross. -- 3rd edition,
pages cm
ISBN 978-1-138-78157-3 (pbk. : alk. paper) - ISBN 978-1-315-76980-6 (ebook)
1. Mechanical engineering—Textbooks. 2. Mechanical engineering—Problems,
exercises, etc. I. Ross, C. T. F., 1935- II. Title.
TJ159.B49 2015
621—dc23
2014024745
ISBN: 9781138781573 (pbk)
ISBN: 9781315769806 (ebk)
Typeset in Times by
Servis Filmsetting Ltd, Stockport, Cheshire
Contents
Preface ix
Part One Revision of Mathematics
i
1 Revisionary mathematics
3
1.1
Introduction
3
1.2
Radians and degrees
4
1.3
Measurement of angles
4
1.4
Triangle calculations
5
1.5
Brackets
8
1.6
Fractions
8
1.7
Percentages
10
1.8
Laws of indices
12
1.9
Simultaneous equations
14
Revision Test 1 Revisionary mathematics
18
2 Further
revisionary mathematics
20
2.1
Units, prefixes and engineering notation
21
2.2
Metric - US/Imperial conversions
24
2.3
Straight line graphs
28
2.4
Gradients, intercepts and equation of a graph 30
2.5
Practical straight line graphs
32
2.6
Introduction to calculus
34
2.7
Basic differentiation revision
34
2.8
Revision of integration
36
2.9
Definite integrals
38
2.10
Simple vector analysis
39
Revision Test 2 Further revisionary mathematics
43
Part Two Statics and Strength
of Materials
45
3 The effects of forces on materials
47
3.1
Introduction
48
3.2
Tensile force
48
3.3
Compressive force
48
3.4
Shear force
48
3.5
Stress
49
3.6
Strain
50
3.7
Elasticity, limit of proportionality
and elastic limit
52
3.8
Hooke’s law
53
3.9 Ductility, brittleness and malleability 57
3.10 Modulus of rigidity 57
3.11 Thermal strain 57
3.12 Compound bars 58
4 Tensile testing 64
4.1 The tensile test 64
4.2 Worked problems on tensile testing 66
4.3 Further worked problems on tensile testing 67
4.4 Proof stress 69
5 Forces acting at a point 71
5.1 Scalar and vector quantities 71
5.2 Centre of gravity and equilibrium 72
5.3 Forces 72
5.4 The resultant of two coplanar forces 73
5.5 Triangle of forces method 74
5.6 The parallelogram of forces method 75
5.7 Resultant of coplanar forces by
calculation 76
5.8 Resultant of more than two coplanar forces 76
5.9 Coplanar forces in equilibrium 78
5.10 Resolution of forces 80
5.11 Summary 83
6 Simply supported beams 86
6.1 The moment of a force 86
6.2 Equilibrium and the principle of moments 87
6.3 Simply supported beams having
point loads 89
6.4 Simply supported beams with couples 93
Revision Test 3 Forces, tensile testing
and beams 97
7 Forces in structures 98
7.1 Introduction 98
7.2 Worked problems on mechanisms
and pin-jointed trusses 99
7.3 Graphical method 100
7.4 Method of joints (a mathematical method) 104
7.5 The method of sections (a mathematical
method) 109
8 Bending moment and shear force diagrams 112
8.1 Bending moment (M) 112
vi Contents
8.2
Shearing force ( F)
113
13.4
Further equations of motion
174
8.3
Worked problems on bending
13.5
Relative velocity
176
moment and shearing force diagrams
113
8.4
Uniformly distributed loads
122
14
Linear momentum and impulse
180
14.1
Linear momentum
180
9 First and second moments of area
127
14.2
Impulse and impulsive forces
183
9.1
Centroids
127
9.2
The first moment of area
128
15
Force,
mass and acceleration
188
9.3
Centroid of area between a curve
15.1
Introduction
188
and the x-axis
128
15.2
Newton’s laws of motion
189
9.4
Centroid of area between a curve and
15.3
Centripetal acceleration
192
the y-axis
128
15.4
Rotation of a rigid body about
9.5
Worked problems on centroids of
a fixed axis
193
simple shapes
129
15.5
Moment of inertia (I)
194
9.6
Further worked problems on centroids
of simple shapes
130
16
Work,
energy and power
197
9.7
Second moments of area of regular
16.1
Work
197
sections
131
16.2
Energy
201
9.8
Second moment of area for ‘built-up’
16.3
Power
202
sections
138
16.4
Potential and kinetic energy
205
16.5
Kinetic energy of rotation
208
Revision Test 4 Forces in structures,
bending moment and shear
Revision Test 6 Linear and angular motion,
force diagrams, and second
momentum and impulse,
moments of area
144
force, mass and acceleration,
work, energy and power
211
10 Bending of beams
145
10.1
Introduction
145
17
Friction
212
a M E
17.1
Introduction to friction
212
10.2
To prove that — = — = —
v I R
146
17.2
Coefficient of friction
213
10.3
Worked problems on the bending
17.3
Applications of friction
214
of beams
147
17.4
Friction on an inclined plane
215
17.5
Motion up a plane with the pulling
11 Torque
151
force P parallel to the plane
215
11.1
Couple and torque
151
17.6
Motion down a plane with the
11.2
Work done and power transmitted
pulling force P parallel to the plane
216
by a constant torque
152
17.7
Motion up a plane due to a horizontal
11.3
Kinetic energy and moment of inertia
154
force P
216
11.4
Power transmission and efficiency
157
17.8
The efficiency of a screw jack
219
12 Twisting of shafts
161
18
Motion in a circle
223
r T GO
18.1
Introduction
223
12.1
To prove that — = — = -
r J L
161
18.2
Motion on a curved banked track
225
12.2
Worked problems on the
18.3
Conical pendulum
226
twisting of shafts
163
18.4
Motion in a vertical circle
228
18.5
Centrifugal clutch
230
Revision Test 5 Bending of beams, torque
and twisting of shafts
167
19
Simple
harmonic motion
232
19.1
Introduction to simple harmonic
motion (SHM)
232
19.2
The spring-mass system
233
Part Three Dynamics
169 ■
19.3
The simple pendulum
235
19.4
The compound pendulum
236
13 Linear and angular motion
171
19.5
Torsional vibrations
237
13.1
The radian
171
13.2
Linear and angular velocity
171
20
Simple
machines
239
13.3
Linear and angular acceleration
173
20.1
Machines
239
Contents vii
20.2
Force ratio, movement ratio
and efficiency
239
20.3
Pulleys
241
20.4
The screw-jack
243
20.5
Gear trains
243
20.6
Levers
245
Revision Test 7 Friction, motion in a circle,
simple harmonic motion and
simple machines 249
Part Four Heat Transfer and Fluid
Mechanics 251
21
Heat energy and transfer
253
21.1
Introduction
253
21.2
The measurement of temperature
254
21.3
Specific heat capacity
255
21.4
Change of state
256
21.5
Latent heats of fusion and vaporisation
257
21.6
A simple refrigerator
259
21.7
Conduction, convection and radiation
259
21.8
Vacuum flask
260
21.9
Use of insulation in conserving fuel
260
22
Thermal expansion
263
22.1
Introduction
263
22.2
Practical applications of thermal
expansion
264
22.3
Expansion and contraction of water
264
22.4
Coefficient of linear expansion
264
22.5
Coefficient of superficial expansion
266
22.6
Coefficient of cubic expansion
267
Revision Test 8 Heat energy and transfer,
and thermal expansion
271
23
Hydrostatics
272
23.1
Pressure
272
23.2
Fluid pressure
274
23.3
Atmospheric pressure
275
23.4
Archimedes’ principle
276
23.5
Measurement of pressure
278
23.6
Barometers
278
23.7
Absolute and gauge pressure
280
23.8
The manometer
280
23.9
The Bourdon pressure gauge
281
23.10
Vacuum gauges
282
23.11
Hydrostatic pressure on submerged
surfaces
282
23.12
Hydrostatic thrust on curved surfaces
284
23.13
Buoyancy
284
23.14
The stability of floating bodies
284
24 Fluid flow 290
24.1
Differential pressure flowmeters
290
24.2
Orifice plate
291
24.3
Venturi tube
292
24.4
Flow nozzle
292
24.5
Pitot-static tube
292
24.6
Mechanical flowmeters
293
24.7
Deflecting vane flowmeter
293
24.8
Turbine type meters
294
24.9
Float and tapered-tube meter
294
24.10
Electromagnetic flowmeter
295
24.11
Hot-wire anemometer
296
24.12
Choice of flowmeter
296
24.13
Equation of continuity
296
24.14
Bernoulli’s equation
297
24.15
Impact of a jet on a stationary plate
298
Ideal gas laws
301
25.1
Boyle’s law
301
25.2
Charles’ law
303
25.3
The pressure or Gay-Lussac’s law
304
25.4
Dalton’s law of partial pressure
305
25.5
Characteristic gas equation
306
25.6
Worked problems on the
characteristic gas equation
306
25.7
Further worked problems on the
characteristic gas equation
308
The measurement of temperature
312
26.1
Liquid-in-glass thermometer
312
26.2
Thermocouples
314
26.3
Resistance thermometers
315
26.4
Thermistors
317
26.5
Pyrometers
317
26.6
Temperature indicating paints
and crayons
319
26.7
Bimetallic thermometers
319
26.8
Mercury-in-steel thermometer
319
26.9
Gas thermometers
319
26.10
Choice of measuring devices
320
Revision Test 9 Hydrostatics, fluid flow,
gas laws and temperature
measurement 322
A list of formulae for mechanical
engineering principles 323
Metric to Imperial conversions and vice versa 328
Greek alphabet 329
Glossary of terms 330
Answers to multiple-choice questions 335
Index 337
This page intentionally left blank
Preface
Mechanical Engineering Principles 3 rd Edition aims to
broaden the reader’s knowledge of the basic principles
that are fundamental to mechanical engineering design
and the operation of mechanical systems.
Modern engineering systems and products still rely
upon static and dynamic principles to make them work.
Even systems that appear to be entirely electronic have a
physical presence governed by the principles of statics.
In this third edition of Mechanical Engineering
Principles , a further chapter has been added on
revisionary mathematics; it is not possible to progress in
engineering studies without a reasonable knowledge of
mathematics, a fact that soon becomes obvious to both
students and teachers alike. It is therefore hoped that this
further chapter on mathematics revision will be helpful
and make engineering studies more comprehensible.
Minor modifications, some further worked problems,
a glossary of terms and famous engineers’ biographies
have all been added to the text.
More has been added to the website for this new edition-
such as full solutions being made available to both stu¬
dents and staff, and much more besides - see page x.
For clarity, the text is divided into four sections, these
being:
Part 1 Revision of Mathematics
Part 2 Statics and Strength of Materials
Part 3 Dynamics
Part 4 Heat Transfer and Fluid Mechanics
Mechanical Engineering Principles 3 rd Edition is
suitable for the following:
(i) National Certificate/Diploma courses in
Mechanical Engineering
(ii) Undergraduate courses in Mechanical,
Civil, Structural, Aeronautical & Marine
Engineering, together with Naval Architecture
(iii) Any introductory/access/foundation course
involving Mechanical Engineering Principles
at University, and Colleges of Further and
Higher education.
Although pre-requisites for the modules covered in this
book include Foundation Certificate/diploma, or similar,
in Mathematics and Science, each topic considered in
the text is presented in a way that assumes that the
reader has little previous knowledge of that topic.
Mechanical Engineering Principles 3 rd Edition
contains over 400 worked problems, followed by over
700 further problems (all with answers). The further
problems are contained within some 150 Exercises;
each Exercise follows on directly from the relevant
section of work, every few pages. In addition, the
text contains 298 multiple-choice questions (all
with answers), and 260 short answer questions,
the answers for which can be determined from the
preceding material in that particular chapter. Where at
all possible, the problems mirror practical situations
found in mechanical engineering. 387 line diagrams
enhance the understanding of the theory.
At regular intervals throughout the text are some
9 Revision Tests to check understanding. For example,
Revision Test 1 covers material contained in Chapter 1,
Test 2 covers the material in Chapter 2, Test 3 covers
the material in Chapters 3 to 6, and so on. No answers
are given for the questions in the Revision Tests, but
an Instructor’s guide has been produced giving full
solutions and suggested marking scheme. The guide is
offered online free to lecturers/instructors - see below.
At the end of the text, a list of relevant formulae is
included for easy reference, together with a glossary
of terms.
‘Learning by Example’ is at the heart of Mechanical
Engineering Principles, 3 rd Edition.
JOHN BIRD
Defence College of Technical Training,
HMS Sultan, formerly
University of Portsmouth and
Highbury College, Portsmouth
CARL ROSS Professor, University of Portsmouth
x Preface
Free Web downloads
The following support material is available
from http://www.routledge.com/cw/bird
For Students:
1. Full worked solutions to all 700 further ques¬
tions contained in the 150 Practice Exercises
2. A list of Essential Formulae
3. A frill glossary of terms
4. Multiple-choice questions
5. Information on 20 Famous Engineers men¬
tioned in the text
6. Video links to practical demonstrations by
Professor Carl Ross http://tiny.cc/6kvqhx
For Lecturers/Instructors:
1-6. As per students 1-6 above.
7. Full solutions and marking scheme for each
of the 9 Revision Tests; also, each test may be
downloaded for distribution to students.
8. All 387 illustrations used in the text may be
downloaded for use in PowerPoint presentations.
Part One
Revision of Mathematics
This page intentionally left blank
Chapter 1
Why it is important to understand: Revisionary mathematics
Mathematics is a vital tool for professional and chartered engineers. It is used in mechanical & manufacturing
engineering, in electrical & electronic engineering, in civil & structural engineering, in naval architecture &
marine engineering and in aeronautical & rocket engineering. In these various branches of engineering, it is
very often much cheaper and safer to design your artefact with the aid of mathematics - rather than through
guesswork. ‘Guesswork’ may be reasonably satisfactory if you are designing an artefact similar to one that
has already proven satisfactory; however, the classification societies will usually require you to provide the
calculations proving that the artefact is safe and sound. Moreover, these calculations may not be readily
available to you and you may have to provide fresh calculations, to prove that your artefact is ‘roadworthy’.
For example, if you design a tall building or a long bridge by ‘guesswork’, and the building or bridge do not
prove to be structurally reliable, it could cost you a fortune to rectify the deficiencies. This cost may dwarf
the initial estimate you made to construct these artefacts, and cause you to go bankrupt. Thus, without
mathematics, the prospective professional or chartered engineer is very severely handicapped.
At the end of this chapter you should be able to:
• convert radians to degrees
• convert degrees to radians
• calculate sine, cosine and tangent for large and small angles
• calculate the sides of a right-angled triangle
• use Pythagoras’theorem
• use the sine and cosine rules for acute-angled triangles
• expand equations containing brackets
• be familiar with summing vulgar fractions
• understand and perform calculations with percentages
• understand and use the laws of indices
• solve simple simultaneous equations
1.1 Introduction
As highlighted above, it is not possible to understand
aspects of mechanical engineering without a good
knowledge of mathematics. This chapter highlights
some areas of mathematics which will make the
understanding of the engineering in the following
chapters a little easier.
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part One
4 Mechanical Engineering Principles
1.2 Radians and degrees
There are hz radians or 360° in a complete circle, thus:
n radians = 180°
from which.
1 rad =
180
o
or
O —
71
71
180
rad
where tt - 3.14159265358979323846 .... to 20 decimal
places!
Convert the following angles to
degrees correct to 3 decimal places:
(a) 0.1 rad (b) 0.2 rad (c) 0.3 rad
180°
(a)
0.1 rad = 0.1 rad x--
ti rad
= 5.730°
o
o
00
t-H
(b)
0.2 rad - 0.2 rad x ,
n rad
= 11.459
o
O
00
t-H
(c)
0.3 rad - 0.3 rad x ,
n rad
= 17.189
o
o
Convert the following angles to
radians correct to 4 decimal places:
(a) 5° (b) 10° (c) 30°
(a) 5° = 5° x = — rad - 0.0873 rad
180° 36
(b) 10° = 10° x = — rad = 0.1745 rad
180° 18
(c) 30° = 30° x = - rad = 0.5236 rad
180° 6
Now try the following Practice Exercise
Practice Exercise 1 Radians and degrees
1. Convert the following angles to degrees
correct to 3 decimal places (where necessary):
(a) 0.6 rad (b) 0.8 rad
(c) 2 rad (d) 3.14159 rad
'(a) 34.377° (b) 45.837°'
(c) 114.592° (d) 180°
2. Convert the following angles to radians
correct to 4 decimal places:
(a) 45° (b) 90°
(c) 120° (d) 180°
(a) — rad or 0.7854 rad
v 7 4
(b) ^-rad or 1.5708 rad
2 71
(c) — rad or 2.0944 rad
3
(d) 7r rad or 3.1416 rad
1.3 Measurement of angles
Angles are measured starting from the horizontal V
axis, in an anticlockwise direction, as shown by 0 l to
0 4 in Figure 1.1. An angle can also be measured in a
clockwise direction, as shown by 0 5 in Figure 1.1, but
in this case the angle has a negative sign before it. If,
for example, 6 4 = 3 00° then 0 5 = -60°.
90°
Figure 1.1
Use a calculator to determine the
cosine, sine and tangent of the following angles,
each measured anticlockwise from the horizontal
6 x 9 axis, each correct to 4 decimal places:
(a) 30° (b) 120° (c) 250°
(d) 320° (e) 390° (f) 480°
(a)
cos 30° = 0.8660
tan 30° - 0.5774
sin 30° = 0.5000
(b)
cos 120° = -0.5000
tan 120° = -1.7321
sin 120° = 0.8660
(c)
cos 250° = - 0.3420
tan 250° = 2.7475
sin 250° =-0.9397
(d)
cos 320° = 0.7660
tan 320° = - 0.8391
sin 320° =-0.6428
Revisionary mathematics 5
(e) cos 390° = 0.8660 sin 390° - 0.5000
tan 390° = 0.5774
(f) cos 480° = - 0.5000 sin 480° = 0.8660
tan 480° = -1.7321
These angles are now drawn in Figure 1.2. Note that
cosine and sine always lie between -1 and +1 but
that tangent can be >1 and <1
Figure 1.2
Note from Figure 1.2 that 6 = 30° is the same as
6 = 390° and so are their cosines, sines and tangents.
Similarly, note that 6 = 120° is the same as 6 = 480°
and so are their cosines, sines and tangents. Also, note
that 6 = -40° is the same as 6 = +320° and so are their
cosines, sines and tangents.
It is noted from above that
• in the first quadrant, i.e. where 0 varies from 0°
to 90°, all (A) values of cosine, sine and tangent are
positive
• in the second quadrant, i.e. where 0 varies from
90° to 180°, only values of sine ( S) are positive
• in the third quadrant, i.e. where 0 varies from
180° to 270°, only values of tangent (7) are positive
• in the fourth quadrant, i.e. where 0 varies from
270° to 360°, only values of cosine (Q are positive
These positive signs, A, S, T and C are shown in
Figure 1.3.
j
.90°
s
A
180°
T
C
270°
Now try the following Practice Exercise
Practice Exercise 2 Measurement of
angles
1. Find the cosine, sine and tangent of the
following angles, where appropriate each
correct to 4 decimal places:
(a)
60°
(b) 90°
(c)
O
o
H
(d)
180°
(e) 210°
00
270°
(g)
330°
(h) -30°
(0
420°
G)
450°
(k) 510°
[(a) 0.5, 0
.8660
1.7321
(b) 0,1, oo
(c) -0.8660, 0.5,-0.5774
(d) -1, 0, 0
(e) -0.8660,-0.5, 0.5774
(f) 0,-1,-oo
(g) 0.8660,-0.5000,-0.5774
(h) 0.8660,-0.5000,-0.5774
(i) 0.5,0.8660,1.7321
G) o, l, co
(k) -0.8660,0.5,-0.5774]
1.4 Triangle calculations
(a) Sine, cosine and tangent
From Figure 1.4, sin 0= — cos 6= —
ac ac
tan 0 =
Figure 1.4
In Figure 1.4, if ab = 2 and ac = 3,
determine the angle 6.
Figure 1.3
It is convenient to use the expression for cos 0 , since
‘ ab ’ and ‘ ac ’ are given.
Part One
Part One
6 Mechanical Engineering Principles
Hence,
from which,
ab 2
cos 6 = ~~ ~ T = 0.66667
LA L J
0 = cos _1 (0.66667) = 48.19°
In Figure 1.4, if be = 1.5 and
ac = 2.2, determine the angle #.
It is convenient to use the expression for sin 0, since
6 be ’ and c ac ’ are given.
be 1.5
Hence, sin 6 = — - t-t- = 0.68182
ac 2.2
from which, 6 = sin _1 (0.68182) = 42.99°
In Figure 1.4, if be = 8 and ab = 1.3,
determine the angle 0.
It is convenient to use the expression for tan 0, since
6 be’ and c ab ’ are given.
Hence,
be 8
tan 0= ~r ~ = 6.1538
ab 1.3
Now try the following Practice Exercise
Practice Exercise 3 Sines, cosines and
tangents and
Pythagoras'theorem
In problems 1 to 5, refer to Figure 1.5.
1. If ab = 2.1 m and be = 1.5 m, determine
angle 6. [35.54°]
2. If ab = 2.3 m and ac = 5.0 m, determine
angle 6. [62.61°]
3. If be = 3.1 m and ac = 6.4 m, determine
angle 0. [28.97°]
4. If ab = 5.7 cm and be = 4.2 cm, determine
the length ac [7.08 cm]
5. If ab = 4.1 m and ac = 6.2 m, determine
length be. [4.65 m]
(c) The sine and cosine rules
For the triangle ABC shown in Figure 1.6,
from which, 9 = tair'^.lSSS) = 80.77°
(b) Pythagoras’ theorem
Pythagoras’ theorem* states that:
(hypotenuse) 2 = (adjacent side) 2 + (opposite side) 2
i.e. in the triangle of Figure 1.5,
ac 2 = ab 2 + be 2
Figure 1.5
In Figure 1.5, if ab = 5.1 m and
be = 6.1 m, determine the length of the
hypotenuse, ac.
From Pythagoras, ac 2 = ab 2 + be 2
= 5.1 2 + 6.7 2 = 26.01 + 44.89
= 70.90
from which, ac = V70.90 = 8.42 m
Revisionary mathematics 7
A
Figure 1.6
the sine rule states:
a
sin A sin B sinC
and the cosine rule states: a 1 = b 2 + c 2 - 2bccos A
Problem In Figure 1.6, if a = 3 m, A = 20° and
B = 120°, determine lengths b, c and angle C.
Using the sine rule,
a
sin^
3
sin B
b
i.e.
from which,
b =
sin 20° sin 120°
3 sin 120° 3x0.8660
sin 20° 0.3420
= 7.596 m
Angle, C = 180° - 20° - 120° - 40°
c a
Using the sine rule again gives:
i.e.
sin C sin A
asinC _ 3 x sin 40°
sin ^4 sin 20°
= 5.638 m
In Figure 1.6, if b = 8.2 cm. c = 5.1 cm
and A = 70°, determine the length a and angles B
and C.
From the cosine rule,
a 2 = b 2 + c 2 - IbccosA
= 8.2 2 + 5.1 2 - 2 x 8.2 x 5.1 x cos70°
= 67.24 + 26.01 - 2(8.2)(5.1)cos70°
= 64.643
Hence, length, a = ^64.643 = 8.04 cm
a b
i.e.
8.04
8.2
sin 70° sin B
from which, 8.04 sin B
and sin B
and
8.2 sin 70°
8.2sin70°
84)4
-l
= 0.95839
B= sin 1 (0.95839) = 73.4P
Since A + B + C = 180°, then
C= m°-A-B= 180°-70°-73.41° = 36.59°
Now try the following Practice Exercise
Practice Exercise 4 Sine and cosine rules
In problems 1 to 4, refer to Figure 1.6.
1. If b = 6 m, c = 4 m and B = 100°, determine
angles A and C and length a.
[A = 38.96°, C = 41.04°, a = 3.83 m]
2. If a = 15 m, c = 23 m and B = 61 °, determine
length b and angles A and C.
[b = 22.01 m, A = 38.86°, C= 74.14°]
3. If a = 4 m, b = 8 m and c = 6 m, determine
angle A. [28.96°]
4. If a = 10.0 cm, b = 8.0 cm and c = 7.0 cm,
determine angles A, B and C.
[A = 83.33°, B = 52.62°, C = 44.05°]
5. In Figure 1.7, PR represents the inclined jib
of a crane and is 10.0 m long. PQ is 4.0 m
long. Determine the inclination of the jib to the
vertical (i.e. angle P) and the length of tie QR.
R
Figure 1.7
Using the sine rule:
sin A sin B
[P= 39.73°, QR = 7.38 m]
Part One
Part One
8 Mechanical Engineering Principles
1.5 Brackets
The use of brackets, which are used in many engi¬
neering equations, is explained through the following
worked problems.
Expand the brackets in problems 3 to 7.
3. 2(x - 2y + 3) [2x-4y + 6]
4. (3x - 4 y) + 3 (y -z)-(z- 4x)
[lx-y-4z\
Expand the bracket to determine A,
given A = a(b + c + d)
Multiplying each term in the bracket by ‘a’ gives:
A = a(b + c + d) = ab + ac + ad
Expand the brackets to determine A,
given A = a[b(c + d) - e(f- g)]
When there is more than one set of brackets the
innermost brackets are multiplied out first. Hence,
A = a[b(c + d) - e(f- g)] = a[bc + bd- ef + eg]
Note that -e x -g = +eg
Now multiplying each term in the square brackets by
6 a ’ gives:
A = abc + abd - aef + aeg
Expand the brackets to determine A ,
given A = a[b(c + d-e) -f(g - h{j - k})]
The inner brackets are determined first, hence
A = a[b{c + d-e)-f(g-h{j- £})]
= a[b(c + d-e) -f(g - hj + hk )]
= a[bc + bd - be -fg + fhj -fhk\
i.e. A = abc + abd - abe - afg + afhj - afhh
Evaluated, given
A = 2[3(6 - 1) - 4(7{2 + 5} - 6)]
A = 2[3(6 - 1) - 4(7{2 + 5} - 6)]
= 2[3(6 - 1) - 4(7 x 7 - 6)]
= 2[3 x 5 - 4 x 43]
= 2[15- 172] = 2[- 157] =-314
Now try the following Practice Exercise
5. 2*+[y-(2x + y)] [0]
6. 24a - [2 {3(5a -b)- 2(a + 2b)} + 3b]
[11/? - 2a]
7. ab[c + d-e(f-g + h{i+j})]
[abc + abd - abef + abeg - abehi - abehj]
1.6 Fractions
An example of a fraction is — where the top line, i.e.
the 2, is referred to as the numerator and the bottom
line, i.e. the 3, is referred to as the denominator.
A proper fraction is one where the numera¬
tor is smaller than the denominator, examples being
2 13 5
, and so on.
3’ 2’ 8’ 16
An improper fraction is one where the denomi¬
nator is smaller than the numerator, examples being
3 2 8 16
—, —, -, —, and so on.
2 13 5
Addition of fractions is demonstrated in the follow¬
ing worked problems.
1 1
Evaluate A, given A = — + -
The lowest common denominator of the two denomi¬
nators 2 and 3 is 6, i.e. 6 is the lowest number that both
2 and 3 will divide into.
13 12. 1 1
Then — = — and - = — i.e. both — and - have the
2 6 3 6 2 3
common denominator, namely 6.
The two fractions can therefore be added as:
1 1 32 3+2 5
A= - + - =- + - =- = -
2 3 6 6 6 6
Practice Exercise 5 Brackets
In problems 1 and 2, evaluate A
1. ,4 = 3(2+ 1+4)
2. A = 4[5(2 + 1) - 3(6 - 7)]
[ 21 ]
[72]
Problem 15.
, . 2 3
Evaluate A, given A = — + —
A common denominator can be obtained by multiply¬
ing the two denominators together, i.e. the common
denominator is 3 x 4 = 12.
Revisionary mathematics 9
The two fractions can now be made equivalent, i.e.
2 8 ,39
= — and = —
3 12 4 12
so that they can be easily added together, as follows:
. 2 3 8 9
A = - + ~ = r +
3 4 12
8 + 9 17
12
12
12
i.e.
2 3 5
A — — + — = 1 —
3 4 12
1 2 3
Evaluate A, given A = — + — + —
A suitable common denominator can be obtained by
multiplying 6x7 = 42, because all three denominators
divide exactly into 42.
2 _ 12
7 _ 42
Thus, — = — ,
6 42
, 3 63
and — = —
2 42
Hence, A
1 2 3
-h-h —
6 7 2
7 12 63
+ — +
7 + 12 + 63 82 41
42 42 42
42
42 21
12 3 ,20
i.e. A = — i - 1 — = 1 —
6 7 2 21
Problem 17.
Determine A as a single fraction,
1
2
given A = — +
—
X
y
A common denominator can be obtained by multiply¬
ing the two denominators together, i.e. xy
Thus, —
x
Hence, A
y
xy
, 2 2x
and — = —
y
i.e.
A =
1 + 2 _ y + 2x
x y xy xy
y + 2 x
xy
Note that addition, subtraction, multiplication and divi¬
sion of fractions may be determined using a calculator
(for example, the CASIO/r-83ES or/x-991ES).
Locate the — and □ — functions on your calculator
(the latter function is a shift function found above
the — function) and then check the following worked
□
problems.
1 2
Evaluate — + —
(i) Press — function
□
(ii) Type in 1
(iii) Press j on the cursor key and type in 4
(iv) i appears on the screen
(v) Press —> on the cursor key and type in +
(vi) Press — function
□
(vii) Type in 2
(viii) Press j on the cursor key and type in 3
(ix) Press —» on the cursor key
(x) Press = and the answer appears
(xi) Press S D function and the fraction changes
to a decimal 0.9166666....
1 2 11
Thus, — + — - — = 0.9167 as a decimal, correct to
4 *3 A £
4 decimal places.
It is also possible to deal with mixed numbers on the
calculator.
Press Shift then the — function and □— appears
□ □
A
Evaluate 5- - 3—
5 4
(i) Press Shift then the — function and □ —
□ □
appears on the screen
(ii) Type in 5 then —> on the cursor key
(iii) Type in 1 and j on the cursor key
(iv) Type in 5 and 5 j appears on the screen
(v) Press —> on the cursor key
(vi) Type in - and then press Shift then the —
□
function and 5 — - □ — appears on the screen
5 □
(vii) Type in 3 then —> on the cursor key
(viii) Type in 3 and j on the cursor key
Part One
Part One
10 Mechanical Engineering Principles
(ix) Type in 4 and 5 — 3 — appears on the screen
5 4
29
(x) Press = and the answer — appears
(xi) Press S <^> D function and the fraction changes
to a decimal 1.45
1 3 29 9
Thus, 5 — - 3— = — = 1— =1.45 as a decimal.
Now try the following Practice Exercise
Practice Exercise 6 Fractions
of commercial life, as well as in engineering. Interest
rates, sale reductions, pay rises, exams and VAT are all
examples where percentages are used.
Percentages are fractions having 100 as their
denominator.
40
For example, the fraction is written as 40% and is
read as ‘forty per cent’.
The easiest way to understand percentages is to go
through some worked examples.
Express 0.275 as a percentage.
0.275 - 0.275 x 100% = 27.5%
In problems 1 to 3, evaluate the given fractions
7
1 1
L 3 + 4
1 1
5 4
1 1 1
3. - +-
6 2 5
12
9
20
7
15
In problems 4 and 5, use a calculator to evaluate
the given expressions
4.
5.
6 .
1 3 8
-x —
3 4 21
3 4 2 4
- X-H- _
4 5 3 9
1
21
9
10
3 5 1
Evaluate — + — - — as a decimal, correct to
= 0.7083
8 9
7. Evaluate 8 — = 2— as a mixed number.
9 3
8 .
3 I
3
117
Evaluate 3-xl—1— as a decimal, cor-
5 3 10
[2.567]
2 3.
9. Determine —l— as a single fraction.
x y
3x + 2 y
xy
rect to 3 decimal places.
1.7 Percentages
Percentages are used to give a common standard. The
use of percentages is very common in many aspects
Express 17.5% as a decimal number.
17.5% =
17.5
Too
= 0.175
Problem 2\
Express
5
8
as a percentage.
1 = ^ x 100% = = 62 . 5 %
o o o
In two successive tests a student
gains marks of 57/79 and 49/67. Is the second mark
better or worse than the first?
57/79 =
57
79
57
— xl00%
79
5700
~79~
= 72.15% correct to 2 decimal places.
49/67 =
49
67
49
— x 100% =
67
4900
~ 67 ~
= 73.13% correct to 2 decimal places.
Hence, the second test is marginally better than the
first test.
This question demonstrates how much easier it is
to compare two fractions when they are expressed as
percentages.
Express 75% as a fraction.
75% =
75
100
3
4
The fraction is reduced to its simplest form
by cancelling, i.e. dividing numerator and denominator
by 25.
Revisionary mathematics 11
Express 37.5% as a fraction.
37.5% =
37.5
Too
375
1000
by multiplying numerator and
denominator by 10
15
40
3
8
by dividing numerator and
denominator by 25
by dividing numerator and
denominator by 5
A drilling speed should be set to
400 rev/min. The nearest speed available on the
machine is 412 rev/min. Calculate the percentage
over-speed.
% over-speed
available speed - correct speed
correct speed
412 - 400 XlOOO/o
xl00%
400
12
400
x 100 % = 3 %
Find 27% of £65.
Now try the following Practice Exercise
27% of £65 =
27
Too
x 65 = £17.55 by calculator
A 160 GB iPod is advertised as
costing £190 excluding VAT. If VAT is added at
20%, what will be the total cost of the iPod?
20
VAT = 20% of £190 = — x 190 =£38
Total cost of iPod = £190 + £38 = £228
A quicker method to determine the total cost is:
1.20 x £190 = £228
Express 23 cm as a percentage of
72 cm, correct to the nearest 1%.
23
23 cm as a percentage of 72 cm = — x 100%
= 31.94444...%
= 32% correct
to the nearest
1 %
A box of screws increases in price
from £45 to £52. Calculate the percentage change
in cost, correct to 3 significant figures.
% change
new value - original value
original value
52 -45 7
x 100 o/ o = _ x 100
x 100 %
45
45
= 15.6% = percentage change in cost
Practice Exercise 7 Percentages
In problems 1 and 2, express the given numbers
as percentages.
1 .
2 .
3.
4.
5.
6 .
7.
8 .
9.
10 .
0.057
0.374
[5.7%]
[37.4%]
Express 20% as a decimal number [0.20]
Express as a percentage [68.75%]
16
Express
13
decimal places
as a percentage, correct to 3
[38.462%]
Place the following in order of size, the
smallest first, expressing each as percent¬
ages, correct to 1 decimal place:
12 9 5
(a) 2l (b) l7 (c) 9 ( d >
'(b) 52.9%,
(c) 55.6%,
Express 65% as a fraction in
form
Calculate 43.6% of 50 kg
Determine 36% of 27 m
6
IT
(d) 54.5%,'
(a) 57.1%.
its simplest
13 '
20 .
[ 21.8 kg]
[9.72 m]
Calculate correct to 4 significant figures:
(a) 18% of 2758 tonnes
(b) 47% of 18.42 grams
(c) 147% of 14.1 seconds
[(a) 496.4 t (b) 8.657 g (c) 20.73 s]
Part One
Part One
12 Mechanical Engineering Principles
11. Express: (a) 140 kg as a percentage of 1 t
(b) 47 s as a percentage of 5 min (c) 13.4 cm
as a percentage of 2.5 m
[(a) 14% (b) 15.67% (c) 5.36%]
12. A computer is advertised on the internet at
£520, exclusive of VAT. If VAT is payable
at 20%, what is the total cost of the com¬
puter? [£624]
13. Express 325 mm as a percentage of 867
mm, correct to 2 decimal places.
[37.49%]
14. When signing a new contract, a Premiership
footballer’s pay increases from £15,500 to
£21,500 per week. Calculate the percentage
pay increase, correct to 3 significant figures.
[38.7%]
15. A metal rod 1.80 m long is heated and its
length expands by 48.6 mm. Calculate the
percentage increase in length. [2.7%]
1.8 Laws of indices
The manipulation of indices, powers and roots is a cru¬
cial underlying skill needed in algebra.
Law 1: When multiplying two or more numbers
having the same base, the indices are added.
For example,
and
More generally,
For example,
2 2 x2 3 = 2 2+3 = 2 5
5 4 x 5 2 x 5 3 = 5 4+2+3 = 5 9
a m x a 11 = a m+n
a 2 x a 4 = a 3+4 = a 1
Law 2: When dividing two numbers having the
same base, the index in the denominator is subtract¬
ed from the index in the numerator.
For example,
and
More generally,
For example,
±-= 2 s - 3 = 2 2
2 3
78
1 _ y 8-5 _ y3
7 5
a
m
a
n
— fltn-n
C 5 2 = C 3
Law 3: When a number which is raised to a power is
raised to a further power, the indices are multiplied.
For example,
2 2 ] _ y2x3 _
and
For example,
3 4 1 = 3 4x2 = 3 8
More generally, ( a m ) n = a mn
d 2 =d 2x3 =d 6
Law 4: When a number has an index of 0, its value is 1.
For example,
and
More generally,
3° = 1
17°= 1
a° = 1
Law 5: A number raised to a negative power is the
reciprocal of that number raised to a positive power.
For example, 3 -4 =
1
and
1
i-3
= 2 3
More generally, a 11 -
a
n
For example, a 2 =
a
Law 6: When a number is raised to a fractional
power the denominator of the fraction is the root
of the number and the numerator is the power.
For example, 8 3 = ^8^ = (2) 2 = 4
and
25 2 = ^25^ =
= ± 5 (Note that = yT)
m
More generally,
For example,
a n = yfa
4
x
m
3 = ^/7
Evaluate in index form 5 3 x 5 x 5 2
5 3 x 5 x 5 2 = 5 3 x 5 1 x 5 2 (Note that 5 means 5 1 )
_ c3+l+2 _
5 6 from law 1
2. Evaluate
Revisionary mathematics 13
From law 2:
= 3 5-4 = 3 1 =3
Evaluate
_ o4-4
from law 2
from law 4
= 2 ° =1
Any number raised to the power of zero equals 1
>4. Evaluate
3x3 :
3 x 3 2 3 1 x 3 2 3 1+2 3 3
3 4 3 4
3 3-4 = 3 - 1
1
3
from laws 1 and 2
from law 5
Evaluate
10 3 x10 2 10 3+2
10 3 x 10 2
10
10
8
10
8
from law 1
10 8 10 8
= 10 5-8 = 10~ 3 from law 2
from law 5
10
+3
1000
10 3 x 10 2
Elence, —^— = 10 ~ 3 =
10
1000
= 0.001
Problem 36. Simplify: (a) (2 3 ) 4 (b) (3 2 ) 5
expressing the answers in index form.
From law 3:
(a) ( 2 3 ) 4 = 2 3x4 = 2 12
(b) (3 2 ) 5 = 3 2x5 = 3 10
Problem 37.
( 10 2 ) 3
Evaluate
N
o
H
X
O
H
From laws 1, 2, and 3:
( 10 2 ) 3 _ 10 (2x3)
10 4 x 10 2 10 (4+2 )
10 6
10 6
= 10 6 " 6
= 10 ° = 1
Evaluate: (a) 4 1/2 (b) 16 3/4
(c) 21 m (d) 9~ m
(a) 4 1/2 = V4 = ±2
(b) 16 3/4 = v/16 3 = ( 2) 3 = 8
(Note that it does not matter whether the 4th root
of 16 is found first or whether 16 cubed is found
first; the same answer will result)
(c) 27 2/3 = a/27 2 = (3 ) 2 = 9
c-i/2 = J_ = J_ = « ± I
9 1/2 79 ±3 3
Simplify a 2 b 3 c x ab 2 c 5
a 2 b 3 c x ab 2 c 5 = a 2 x b 3 x c x a x b 2 x c 5
= a 2 x b 3 x c l x a 1 x b 2 x c 5
Grouping together like terms gives:
a 2 x a [ x b 3 x b 2 x c 1 x c 5
Using law 1 of indices gives:
a 2 +i x £3+2 x c l+5 = a 3 X b 5 x c 6
i.e. a 2 b 3 c x ab 2 c s = a 3 b 5 c 6
Simplify
5 2
x y z
2 3
x yz
<; 0 S')
x y z _ x x y x z
17 . 9 7
x yz x X y x z
= x 5 2 x y 2 1 x z
5 2
x y z
— x — x —
x
1-3
y
i
by law 2
x 3 x y l x z 2
x 3 yz 2
or
x° y
Now try the following Practice Exercise
Practice Exercise 8 Laws of indices
In questions 1 to 18, evaluate without the aid of
a calculator
1. Evaluate 2 2 x 2 x 2 4 [ 2 7 = 128]
2. Evaluate 3 5 x 3 3 x 3 in index form
3.
4.
5.
Evaluate
Evaluate
7
2 3
3 3
Evaluate 7°
[3 9 ]
[2 4 =16]
[ 1 ]
Part One
Part One
14 Mechanical Engineering Principles
6 .
7.
8 .
9.
10 .
11 .
Evaluate
2 j x 2 x 2 b
[ 2 3 = 8 ]
2 7
Evaluate
10 xl0 6
[10 2 = 100]
10 -
Evaluate
10 4 +
10
O
o
o
H
II
m
O
i—H
Evaluate
10 3 xio 4
10
9
io- 2 - 1 -
10 2
1 -o.oi
100
Evaluate
5 6 x5 2 4- 5 7
[5]
Evaluate (7 2 ) 3 in index form
[7 6 ]
31.
32.
1
/
\4
U 8 1
[c 14 ]
or b 12
12 .
Evaluate (3 3 ) 2
[3 6 =
-729]
13.
Evaluate
3 7 x 3 4
•-/ •_/ • • -t p
in innPY tnrm
[3 6 ]
3 5
14.
Evaluate
(9x3 ‘ )3 in index form
(3 x 27) 2
[3 4 ]
15.
Evaluate
(16 x4 ) 2
(2 x 8) 3
[ 1 ]
16.
Evaluate
5~ 2
5~ 4
[5 2
= 25]
17.
Evaluate
3 2 x3~ 4
3 -5 - 1 -
1
3 3
i _ 3 5 _
243
18.
Evaluate
7 2 x 7 -3
[7 2
= 49]
7x7~ 4
In problems 19 to 36, simplify the following,
giving each answer as a power
19.
Z 2 X z 6
[z 8 ]
20 .
ax a 2 x a 5
[fl 8 ]
21 .
/ 2 8 x n~ 5
[« 3 ]
22 .
b 4 xb 7
[b n ]
23.
b 2 + b 5
b~ 3 or \
b 3
24.
25.
C 3 X C 3 -i- c 4
m 5 x m 6
[c 4 ]
[m 4 ]
m 4 x m?
26.
(x 2 )(x)
-3 1 '
x J or
[Z> 10 ]
or s
35. p 2 qr 2 x p 2 q 5 r x pqr 2
[p 6 q 7 r 5 \
36.
3 2
x y z
~~5 3
xyz
or
y
2 2
X Z
1.9 Simultaneous equations
The solution of simultaneous equations is demonstrat¬
ed in the following worked problems.
If 6 apples and 2 pears cost £1.80
and 8 apples and 6 pears cost £2.90, calculate how
much an apple and a pear each cost.
Let an apple = A and a pear = P, then:
6A + 2P= 180
8 A + 6P = 290
From equation (1 ), 6A = 180-2 P
and A = - = 30 - 0.3333P
6
From equation (2), 8 A = 290 - 6P
and A = 290 Z 6P . = 36.25 - 0.75 P
8
Equating (3) and (4) gives:
30 - 0.3333P = 36.25 - 0.75P
(1)
( 2 )
(3)
(4)
Revisionary mathematics 15
i.e. 0.75P - 0.3333 P = 36.25 - 30
and 0.4167P = 6.25
and
n — 6.25
0.4167
Substituting in (3) gives: A = 30- 0.3333(15)
= 30-5=25
Hence, an apple costs 25p and a pear costs 15p
The above method of solving simultaneous equations
is called the substitution method.
If 6 bananas and 5 peaches cost
£3.45 and 4 bananas and 8 peaches cost £4.40,
calculate how much a banana and a peach each
cost.
Let a banana = B and a peach = P, then:
6B + 5P = 345
(i)
4B + 8P = 440
(2)
Multiplying equation (1) by 2 gives:
\2B + 10P = 690
(3)
Multiplying equation (2) by 3 gives:
12P + 24P = 1320
(4)
Equation (4) - equation (3) gives: 14P = 630
from which,
P=^=45
14
Substituting in (1) gives: 6B + 5(45) = 345
i.e.
i.e.
6B = 345 - 5(45)
6B= 120
and
*=!S-20
Hence, a banana costs 20p and a peach costs 45p
The above method of solving simultaneous equations
is called the elimination method.
If 20 bolts and 2 spanners cost £10,
and 6 spanners and 12 bolts cost £18, how much
does a spanner and a bolt cost?
and 6s + \2b =18 (2)
Multiplying equation (1) by 3 gives:
6s + 60 b = 30 (3)
Equation (3) - equation (2) gives: 48 b = 12
from which,
b = = 0.25
48
Substituting in (1) gives: 2s + 20(0.25) = 10
i.e.
2s = 10-20(0.25)
i.e.
2s = 5
and
s= J =2.5
2
Therefore, a spanner costs £2.50 and a bolt costs
£0.25 or 25p
Now try the following Practice Exercises
Practice Exercise 9 Simultaneous
equations
1. If 5 apples and 3 bananas cost £1.45 and
4 apples and 6 bananas cost £2.42, deter¬
mine how much an apple and a banana each
cost. [apple = 8p, banana = 35p]
2. If 7 apples and 4 oranges cost £2.64 and
3 apples and 3 oranges cost £1.35, determine
how much an apple and an orange each cost.
[apple = 28p, orange = 17p]
3. Three new cars and four new vans sup¬
plied to a dealer together cost £93000, and
five new cars and two new vans of the same
models cost £99000. Find the respective
costs of a car and a van.
[car = £15000, van = £12000]
4. In a system of forces, the relationship
between two forces F { and F 2 is given by:
5Fj + 3F 2 = -6
3F l + 5F 2 = - 18
Solve for F { and F 2
[Fj = 1.5, F 2 = -4.5]
5.
Let
Therefore,
s = a spanner and b = a bolt.
2s + 20b = 10
(i)
Solve the simultaneous equations:
a + b = 7
a-b = 3 [a = 5, b = 2]
Part One
Part One
16 Mechanical Engineering Principles
6. Solve the simultaneous equations:
8^-36 = 51
3a + 4b= 14 [a = 6,b = - 1]
Practice Exercise 10 Multiple-choice
questions on
revisionary
mathematics
(Answers on page 335)
1. 73° is equivalent to:
(a) 23.24 rad (b) 1.274 rad
(c) 0.406 rad (d) 4183 rad
2. 0.52 radians is equivalent to:
(a) 93.6° (b) 0.0091°
(c) 1.63° (d) 29.79°
3. 37t/ 4 radians is equivalent to:
(a) 135° (b) 270°
(c) 45° (d) 67.5°
4. In the right-angled triangle ABC shown in
Figure 1.8, sinev4 is given by:
(a) bla (b) db
(c) blc (d) alb
Figure 1.8
5. In the right-angled triangle ABC shown in
Figure 1.8, cosine C is given by:
(a) alb (b) db
(c) ale (d) bla
6. In the right-angled triangle ABC shown in
Figure 1.8, tangents is given by:
(a) blc (b) ale
(c) alb (d) da
7. In the right-angled triangle PQR shown in
Figure 1.9, angle R is equal to:
(a) 41.41° (b) 48.59°
(c) 36.87° (d) 53.13°
Figure 1.9
8. In the triangle ABC shown in Figure 1.10,
side V is equal to:
(a) 61.27 mm
(b) 86.58 mm
(c) 96.41 mm
(d) 54.58 mm
A
Figure 1.10
9. In the triangle ABC shown in Figure 1.10,
angle B is equal to:
(a) 0.386° (b) 22.69°
(c) 74.71° (d) 23.58°
10. Removing the brackets from the expres¬
sion: a[b + 2 c-d{(e-f)-g(m-n)} \ gives:
(a) ab + 2 ac - ade - adf+ adgm - adgn
(b) ab + 2 ac - ade - adf— adgm - adgn
(c) ab + lac - ade + adf + adgm - adgn
(d) ab + lac - ade - adf + adgm + adgn
n 5 i 2 . 1+
11. — +-is equal to:
6 5 3
(a) I (b)
(0-1 mijl
Revisionary mathematics 17
- is equal to:
3
is equal to:
T , • • ■ (16 x4) 2 .
The engineering expression -— is
, . (8 x 2) 4
equal to:
(a) 4 (b) 2-‘
(■=) ») 1
_2
(16 4_27 3 ) is equal to:
(•) ^ (b) - 7
(C) if (d) -sT
14. 11 mm expressed as a percentage of
41 mm is:
(a) 2.68, correct to 3 significant figures
(b) 2.6, correct to 2 significant figures
(c) 26.83, correct to 2 decimal places
(d) 0.2682, correct to 4 decimal places
2 -3
The value of-- 1 is equal to:
2" 4
(a) 1 (b) 2
(c) ~ (d) T
In an engineering equation
value of r is:
(a) - 6 (b) 2
(c) 6 (d) - 2
I. The
9
_ 3
16 4 is equal to:
(a) 8 (b) ~
2
(C) 4 (d) I
20. The solution of the simultaneous equa¬
tions: 3a - 2b = 13 and 2a + 5b = - 4
is:
(a) a = - 2, b = 3
(b) a = 1, b = - 5
(c) a = 3, b =-2
(d) a = - 7, b = 2
References
There are many aspects of mathematics needed in engineering
studies; a few have been covered in this chapter. For
further engineering mathematics, see the following
references:
[1] BIRD J. O. Basic Engineering Mathematics 6th Edition ,
Taylor & Francis, 2014
[2] BIRD J. O. Engineering Mathematics 7 th Edition, Taylor &
Francis, 2014
For fully worked solutions to each of the problems in Practice Exercises 1 to 10 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part One
Part One
Revision Test 7 Revisionary mathematics
This Revision Test covers the material contained in Chapter 1. The marks for each question are shown in brackets
at the end of each question.
1. Convert, correct to 2 decimal places:
(a) 76.8° to radians
(b) 1.724 radians to degrees (4)
2. In triangle JKL in Figure RT1.1, find:
(a) length KJ correct to 3 significant figures
(b) sin L and tan K , each correct to 3 decimal places
K
L
Figure RT1.1
(4)
3. In triangle PQR in Figure RT1.2, find angle P in
decimal form, correct to 2 decimal places
R
Figure RT1.2
( 2 )
4. In triangle ABC in Figure RT1.3, find lengths AB
and AC, correct to 2 decimal places
4
Figure RT1.3
(4)
5. A triangular plot of land ABC is shown in
Figure RT1.4. Solve the triangle and determine its
area
A
Figure RT1.4
(9)
6 .
Figure RT1.5 shows a roof truss PQR with rafter
PQ = 3 m. Calculate the length of (a) the roof rise
PP' (b) rafter PR, and (c) the roof span QR. Find
also (d) the cross-sectional area of the roof truss
Figure RT1.5
( 10 )
7.
8 .
9.
10 .
11 .
Solve triangle ABC given b — 10 cm, c = 15 cm
and ZZ = 60°. (7)
Remove the brackets and simplify
2(3x - 2 y) - (4 y - 3x)
Remove the brackets and simplify
10a - [3(2a - b) - 4(b -a) + 5b]
(3)
(4)
Determine, correct to 2 decimal places, 57% of
17.64 g (2)
Express 54.7 mm as a percentage of 1.15 m,
correct to 3 significant figures. (3)
12. Simplify:
, x 3 7
(a)-
4 15
(b) 1— — 2— + 3 —
8 3 6
( 8 )
Revisionary mathematics 19
13. U se a calculator to evaluate:
7 3 3
(a) l-x-x3-
9 8 5
(c) 1-X2- + -
3 5 5
14. Evaluate:
(a) 3 x 2 3 x 2 2
15.
16.
(b) 492
Evaluate:
( 10 )
( 4 )
(a)
7
(b)
2 2
Evaluate:
. . 2 3 x 2 x 2 2
(a) -;-
10 4 x 10 x 10 5
10 6 x 10 2
( 4 )
(b)
(c)
2 3 x 16
(8 x 2 ) ;
'! V1
4 2
y
17. Evaluate:
/ \
-2
(a) (27) 4 (b)
, 3 y
2
9
18. Solve the simultaneous equations:
(a) 2 x + y = 6
5x-y = 22
(b) 4x-3y=ll
3x + 5y = 30
( 7 )
( 5 )
( 10 )
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 1,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
Part One
Chapter 2
Further revisionary
mathematics
Why it is important to understand: Further revisionary mathematics
In engineering there are many different quantities to get used to, and hence many units to become familiar
with. For example, force is measured in newtons, electric current is measured in amperes and pressure is
measured in pascals. Sometimes the units of these quantities are either very large or very small and hence
prefixes are used. For example, 1000 pascals may be written as 10 3 Pa which is written as 1 kPa in prefix
form, the k being accepted as a symbol to represent 1000 or 10 3 . Studying, or working, in an engineering
discipline, you very quickly become familiar with the standard units of measurement, the prefixes used and
engineering notation. An electronic calculator is extremely helpful with engineering notation.
Most countries have used the metric system of units for many years; however, there are other coun¬
tries, such as the USA, who still use the imperial system. Hence, metric to imperial unit conversions,
and vice-versa, are internationally important and are contained in this chapter.
Graphs have a wide range of applications in engineering and in physical sciences because of their
inherent simplicity. A graph can be used to represent almost any physical situation involving discrete
objects and the relationship among them. If two quantities are directly proportional and one is plotted
against the other, a straight line is produced. Examples of this include an applied force on the end of a
spring plotted against spring extension, the speed of a flywheel plotted against time, and strain in a wire
plotted against stress (Hooke’s law). In engineering, the straight line graph is the most basic graph to
draw and evaluate.
There are many practical situations engineers have to analyse which involve quantities that are vary¬
ing. Typical examples include the stress in a loaded beam, the temperature of an industrial chemical, the
rate at which the speed of a vehicle is increasing or decreasing, the current in an electrical circuit or the
torque on a turbine blade. Differential calculus, or differentiation, is a mathematical technique for analys¬
ing the way in which functions change. This chapter explains how to differentiate the five most common
functions. Engineering is all about problem solving and many problems in engineering can be solved using
calculus. Physicists, chemists, engineers, and many other scientific and technical specialists use calculus in
their everyday work; it is a technique of fundamental importance. Integration has numerous applications
in engineering and science and some typical examples include determining areas, mean and r.m.s. values,
volumes of solids of revolution, centroids, second moments of area and differential equations. Standard
integrals are covered in this chapter, and for any further studies in engineering, differential and integral
calculus are unavoidable.
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Further revisionary mathematics 21
Vectors are an important part of the language of science, mathematics, and engineering. They are
used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in
structures and materials, and flows of atmospheres and fluids, and they have many other applications.
Resolving a vector into components is a precursor to computing things with or about a vector quantity.
Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quan¬
tities, resolving vectors into components is a most important skill required in any engineering studies.
At the end of this chapter, you should be able to:
• state the seven SI units
• understand derived units
• recognise common engineering units
• understand common prefixes used in engineering
• use engineering notation and prefix form with engineering units
• understand and calculate metric to imperial conversions and vice versa
• understand rectangular axes, scales and co-ordinates
• plot given co-ordinates and draw the best straight line graph
• determine the gradient and vertical-axis intercept of a straight line graph
• state the equation of a straight line graph
• plot straight line graphs involving practical engineering examples
• state that calculus comprises two parts - differential and integral calculus
• differentiate y = ax' 1 by the general rule
• differentiate sine, cosine, exponential and logarithmic functions
• understand that integration is the reverse process of differentiation
• determine integrals of the form ax n where n is fractional, zero, or a positive or negative integer
• integrate standard functions - cos ax, sin ax, e ax , J_
• evaluate definite integrals x
• evaluate 2 by 2 and 3 by 3 determinants
• determine scalar (or dot) products of two vectors
• determine vector (or cross) products of two vectors
2.1 Units, prefixes and engineering
notation
Of considerable importance in engineering is knowl¬
edge of units of engineering quantities, the prefixes
used with units, and engineering notation.
We need to know, for example, that
80 kN = 80 x 10 3 N which means 80,000
newtons
and 25 mJ = 25 x 10 _3 J which means 0.025 joules
and 50 nF = 50 x 10 _9 F which means 0.000000050
farads
This is explained in this chapter.
SI units
The system of units used in engineering and science
is the Systeme Internationale d’Unites (International
System of Units), usually abbreviated to SI units, and
is based on the metric system. This was introduced in
1960 and is now adopted by the majority of countries
as the official system of measurement.
The basic seven units used in the SI system are listed
on page 22 with their symbols.
There are, of course, many other units than the seven
shown. These other units are called derived units and
are defined in terms of the standard units listed.
For example, speed is measured in metres per second,
therefore using two of the standard units, i.e. metres
and seconds.
Part One
Part One
22 Mechanical Engineering Principles
Quantity
Unit
Symbol
Length
metre
m
(1 m = 100 cm
= 1000 mm)
Mass
kilogram
kg
(1kg = 1000 g)
Time
second
s
Electric
current
ampere
A
Thermo¬
dynamic
temperature
kelvin
K
(K = °C + 273)
Luminous
intensity
candela
cd
Amount of
substance
mole
mol
Some derived units are given special names.
For example, force = mass x acceleration, has units of
kilogram metre per second squared, which uses three of
the base units, i.e. kilograms, metres and seconds. The
unit of kg m/s 2 is given the special name of a newton.
Below is a list of some quantities and their units that
are common in engineering.
Quantity
Unit
Symbol
Length
metre
m
Area
square metre
2
m
Volume
cubic metre
m 3
Mass
kilogram
kg
Time
second
s
Electric current
ampere
A
Speed, velocity
metre per second
m/s
Acceleration
metre per second
squared
m/s 2
Density
kilogram per cubic
metre
kg/m 3
Temperature
kelvin or Celsius
K or °C
Angle
radian or degree
rad or °
Angular velocity
radian per second
rad/s
Frequency
hertz
Hz
Force
newton
N
Pressure
pascal
Pa
Energy, work
joule
J
Power
watt
W
Electric potential
volt
V
Capacitance
farad
F
Electrical
resistance
ohm
Q
Inductance
henry
H
Moment of force
newton metre
Nm
Stress
pascal
Pa
Torque
newton metre
N m
Momentum
kilogram metre per
second
kg m/s
Common prefixes
SI units may be made larger or smaller by using pre¬
fixes which denote multiplication or division by a par¬
ticular amount.
The most common multiples are listed below. Knowl¬
edge of indices is needed since all of the prefixes are
powers of 10 with indices that are a multiple of 3.
Prefix
Name
Meaning
T
tera
multiply
by 10 12
i.e. x 1,000,000,000,000
G
giga
multiply
by 10 9
i.e. x 1,000,000,000
M
mega
multiply
by 10 6
i.e. x 1,000,000
k
kilo
multiply
by 10 3
i.e. x 1,000
m
milli
multiply
by 10“ 3
i.e. x 1 = 1 =0.001
10 3 1000
micro
multiply
by 10“ 6
1 1
1 £ V _
10 6 ~ 1,000,000
= 0.000001
n
nano
multiply
by 10- 9
1 1
Le ' X 10 9 - 1,000,000,000
= 0.000 000 001
P
pico
multiply
by 10- 12
1 1
l e x 10 12 - ^000,000,000,000
= 0.000 000 000 001
Here are some examples of prefixes used with engi¬
neering units.
A frequency of 15 GHz means 15 x 10 9 Hz, which
is 15,000,000,000 hertz*, i.e. 15 gigahertz is written
as 15 GHz and is equal to 15 thousand million hertz.
Instead of writing 15,000,000,000 hertz, it is much
neater, takes up less space and prevents errors caused
by having so many zeros, to write the frequency as
15 GHz.
Further revisionary mathematics 23
A voltage of 40 MV means 40 x 10 6 V, which is
40,000,000 volts, i.e. 40 megavolts is written as 40 MV
and is equal to 40 million volts.
12 12
Energy of 12 mJ means 12 x 10 -3 J or — J or-J,
10 ' 1000
which is 0.012 J, i.e. 12 millijoules is written as 12 mJ
and is equal to 12 thousandths of a joule.
A time of 150 ns means 150 x 10~ 9 s or s, which is
i ° 9
0.000 000 150 s, i.e. 150 nanoseconds is written as 150 ns
and is equal to 150 thousand millionths of a second.
A force of 20 kN means 20 x 10 3 N, which is 20,000
newtons, i.e. 20 kilonewtons is written as 20 kN and is
equal to 20 thousand newtons.
Engineering notation
Engineering notation is a number multiplied by a power
of 10 that is always a multiple of 3.
For example,
43,645 = 43.645 x 10 in engineering notation and
0.0534 = 53.4x10”' in engineering notation
In the list of engineering prefixes on page 22 it is ap¬
parent that all prefixes involve powers of 10 that are
multiples of 3.
* Heinrich Rudolf Hertz (22 February 1857-1 January 1894)
was the first person to prove the existence of electromagnetic
waves. The scientific unit of frequency was named hertz in his
honour. To find out more go to www.routledge.com/cw/bird
For example, a force of 43,645 N can be re-written as
43.645 x 10 3 N and from the list of prefixes can then be
expressed as 43.645 kN. Thus,
43,645 N = 43.645 kN
To help further, on your calculator is an ‘ENG’ button.
Enter the number 43,645 into your calculator and then
press ‘=’
Now press the ‘ENG’ button and the answer is
43.645x10'.
We then have to appreciate that 10 3 is the prefix ‘kilo’
giving 43,645 N = 43.645 kN
In another example, let a current be 0.0745 A
Enter 0.0745 into your calculator. Press ‘=’
Now press ‘ENG’ and the answer is 74.5 x 10”'.
We then have to appreciate that 10 -3 is the prefix ‘milli’
giving 0.0745 A = 74.5 mA
Express the following in engineering
notation and in prefix form:
(a) 300,000 W (b) 0.000068 H
(a) Enter 300,000 into the calculator. Press ‘=’
Now press ‘ENG’ and the answer is 300 x 10 3
From the table of prefixes on page 22, 10 3 cor¬
responds to kilo.
Hence, 300,000 W = 300 x 10 3 W in engineering
notation
= 300 kW in prefix form
(b) Enter 0.000068 into the calculator. Press ‘=’
Now press ‘ENG’ and the answer is 68 x 10 -6
From the table of prefixes on page 22, 10~ 6 cor¬
responds to micro.
Hence, 0.000068 H = 68 x 10 -6 H in engineering
notation
= 68 pH in prefix form
Express the following in engineering
notation and in prefix form:
(a) 42 x 10 5 Q (b) 47 x 10- 10 F
(a) Enter 42 x 10 5 into the calculator. Press ‘=’
Now press ‘ENG’ and the answer is 4.2 x 10 6
From the table of prefixes on page 22, 10 6 cor¬
responds to mega.
Part One
Part One
24 Mechanical Engineering Principles
Hence, 42 x 10 5 Q = 4.2 x 10 6 Q in engineering
notation
= 4.2 MQ in prefix form
47
(b) Enter 47 -s- 10 10 = - into the cal-
culator. Press <=’ 10,000,000,000
Now press ‘ENG’ and the answer is 4.7 x 10 -9
From the table of prefixes on page 22, 10 -9 cor¬
responds to nano.
Hence, 47 -s- 10 _10 F = 4.7 x 10~ 9 F in engineering
notation
= 4.7 nF in prefix form
Rewrite (a) 14,700 mm in metres
(b) 276 cm in metres (c) 3.375 kg in grams
(a) 1 m = 1000 mm
Hence, 1 mm =-= —- = 10 3 m
1000 10 3
Hence, 14,700 mm = 14,700 x 10 -3 m = 14.7 m
1 1
(b) 1 m = 100 cm hence 1 cm = -= —- = 10 m
100 10 2
Hence, 276 cm = 276 x 10 -2 m = 2.76 m
(c) 1 kg = 1000 g = 10 3 g
Hence, 3.375 kg - 3.375 x 10 3 g = 3375 g
Now try the following Practice Exercise
Practice Exercise 11 Engineering notation
In problems 1 to 12, express in engineering nota-
tion in prefix form:
1. 60,000 Pa
[60 kPa]
2 .
0.00015 W
[150 pW or 0.15 mW]
3.
5 x 10 7 V
[50 MV]
4.
5.5 x 10- 8 F
[55 nF]
5.
100,000 N
[100 kN]
6 .
0.00054 A
[0.54 mA or 540 pA]
7.
15 x 10 s Q.
[1.5 MQ]
8 .
225 x 10 " 4 V
[22.5 mV]
9.
35,000,000,000 Hz
[35 GHz]
10 .
1.5 x 10 " 11 F
[15 P F]
11 .
0.000017 A
[17 pA]
12 .
46200 Q
[46.2 kO]
13.
Rewrite 0.003 mA in pA
[3 pA]
14.
Rewrite 2025 kHz as MHz
[2.025 MHz]
15.
Rewrite 5 x 10 4 N in kN
[50 kN]
16.
Rewrite 300 pF in nF
[0.3 nF]
17.
Rewrite 6250 cm in metres
[62.50 m]
18.
Rewrite 34.6 g in kg
[0.0346 kg]
19.
The tensile stress acting on
a rod is 5600000
Pa. Write this value in engineering notation.
[5.6 x 10 6 Pa = 5.6 MPa]
20. The expansion of a rod is 0.0043 m. Write
this in engineering notation.
[4.3 x 10 -3 m = 4.3 mm]
2.2 Metric - US/Imperial Conversions
The Imperial System (which uses yards, feet, inches,
etc. to measure length) was developed over hundreds of
years in the UK, then the French developed the Metric
System (metres) in 1670, which soon spread through
Europe, even to England itself in 1960. But the USA
and a few other countries still prefer feet and inches.
When converting from metric to imperial units,
or vice versa, one of the following tables ( 2.1 to 2 . 8 )
should help.
Table 2.1 Metric to Imperial length
Metric
US or Imperial
1 millimetre, mm
0.03937 inch
1 centimetre, cm = 10 mm
0.3937 inch
1 metre, m = 100 cm
|l.0936
1 kilometre, km = 1000 m
0.6214 mile
Calculate the number of inches in
350 mm, correct to 2 decimal places.
350 mm = 350 x 0.03937 inches = 13.78 inches from
Table 2.1
Further revisionary mathematics 25
Calculate the number of inches in
52 cm, correct to 4 significant figures.
52 cm = 52 x 0.3937 inches = 20.47 inches from Table 2.1
Calculate the number of yards in
74 m, correct to 2 decimal places.
74 m = 74 x 1.0936 yards = 80.93 yds from Table 2.1
Calculate the number of miles in
12.5 km, correct to 3 significant figures.
12.5 km = 12.5 x 0.6214 miles = 7.77 miles from
Table 2.1
Table 2.2 Imperial to Metric length
US or Imperial
Metric
1 inch, in
2.54 cm
1 foot, ft = 12 in
0.3048 m
1 yard, yd = 3 ft
0.9144 m
1 mile = 1760 yd
1.6093 km
1 nautical mile = 2025.4 yd
1.852 km
Calculate the number of centimetres
in 35 inches, correct to 1 decimal place.
35 inches = 35 x 2.54 cm = 88.9 cm from Table 2.2
Calculate the number of metres in 66
inches, correct to 2 decimal places.
66 inches = — feet = — x 0.3048 m = 1.68 m from
12 12 Table 2.2
Calculate the number of metres in
50 yards, correct to 2 decimal places.
50 yards = 50 x 0.9144 m = 45.72 m from Table 2.2
Calculate the number of kilometres
in 7.2 miles, correct to 2 decimal places.
7.2 miles = 7.2 x 1.6093 km = 11.59 km from Table 2.2
Calculate the number of (a) yards
(b) kilometres in 5.2 nautical miles.
(a) 5.2 nautical miles = 5.2 x 2025.4 yards
= 10532 yards from Table 2.2
(b) 5.2 nautical miles = 5.2 x 1.852 km = 9.630 km
from Table 2.2
Table 2.3 Metric to Imperial area
Metric
US or Imperial
1 cm 2 =100 mm 2
0.1550 in 2
1 m 2 = 10,000 cm 2
1.1960 yd 2
1 hectare, ha = 10,000 m 2
2.4711 acres
1 km 2 = 100 ha
0.3861 mile 2
Calculate the number of square
inches in 47cm 2 , correct to 4 significant figures.
47cm 2 = 47 x 0.1550 in 2 = 7.285 in 2 from Table 2.3
Calculate the number of square
yards in 20 m 2 , correct to 2 decimal places.
20 m 2 = 20 x 1.1960 yd 2 = 23.92 yd 2 from Table 2.3
Calculate the number of acres in 23
hectares of land, correct to 2 decimal places.
23 hectares = 23 x 2.4711 acres = 56.84 acres from
Table 2.3
Calculate the number of square miles
in a field of 15 km 2 area, correct to 2 decimal places.
15 km 2 = 15 x 0.3861 mile 2 = 5.79 mile 2 from Table 2.3
Table 2.4 Imperial to Metric area
US or Imperial
Metric
1 in 2
6.4516 cm 2
1 ft 2 = 144 in 2
0.0929 m 2
1 yd 2 = 9 ft 2
0.8361m 2
1 acre = 4840 yd 2
4046.9 m 2
1 mile 2 = 640 acres
2.59 km 2
Calculate the number of square cen¬
timetres in 17.5 in 2 , correct to the nearest square
centimetre.
17.5 in 2 = 17.5 x 6.4516 cm 2 = 113 cm 2 from Table 2.4
Calculate the number of square
metres in 205 ft 2 , correct to 2 decimal places.
205 ft 2 = 205 x 0.0929 m 2 = 19.04 m 2 from Table 2.4
Part One
Part One
26 Mechanical Engineering Principles
Calculate the number of square metres
in 11.2 acres, correct to the nearest square metre.
11.2 acres =
11.2 x 4046.9 m 2 = 45325 m 2 from
Table 2.4
Problem 20.
Calculate the number of square
kilometres in
12.6 mile 2 , correct to 2 decimal places.
12.6 mile 2 , =
12.6 x 2.59 km 2 = 32.63 km 2 from
Table 2.4
Table 2.5 Metric to Imperial volume/capacity
Metric
US or Imperial
1 cm 3
0.0610 in 3
1 dm 3 = 1000 cm 3 0.0353 ft 3
1 m 3 = 1000 dm 3 1.3080 yd 3
1 litre = 1 dm 3 = 1000 cm 3 2.113 fluid pt= 1.7598 pt
Calculate the number of cubic
inches in 123.5 cm 3 , correct to 2 decimal places.
123.5cm 3 - 123.5 x 0.0610 in 3 = 7.53 in 3 from
Table 2.5
Calculate the number of cubic feet
in 144 dm 3 , correct to 3 decimal places.
144 dm 3 - 144 x 0.0353 ft 3 - 5.083 ft 3 from Table 2.5
Calculate the number of cubic yards
in 5.75 m 3 , correct to 4 significant figures.
5.75 m 3 = 5.75 x 1.3080 yd 3 - 7.521 yd 3 from Table 2.5
Calculate the number of US fluid
pints in 6.34 litres of oil, correct to 1 decimal place.
6.34 litre = 6.34 x 2.113 US fluid pints = 13.4 US fluid
pints from Table 2.5
Table 2.6 Imperial to Metric volume/capacity
US or Imperial
Metric
1 in 3
16.387 cm 3
1 ft 3
0.02832 m 3
1 US floz= 1.0408 UK floz
0.0296 litres
1 US pint (16 fl oz) = 0.8327 UK pt
0.4732 litres
1 US gal (231 in 3 ) = 0.8327 UK gal
3.7854 litres
Calculate the number of cubic cen¬
timetres in 3.75 in 3 , correct to 2 decimal places.
3.75 in 3 = 3.75 x 16.387 cm 3 = 61.45 cm 3 from Table 2.6
Calculate the number of cubic me¬
tres in 210 ft 3 , correct to 3 significant figures.
210 ft 3 = 210 x 0.02832 m 3 = 5.95 m 3 from Table 2.6
Calculate the number of litres in
4.32 US pints, correct to 3 decimal places.
4.32 US pints = 4.32 x 0.4732 litres = 2.044 litres from
Table 2.6
Calculate the number of litres in
8.62 US gallons, correct to 2 decimal places.
8.62 US gallons = 8.62 x 3.7854 litres = 32.63 litres
from Table 2.6
Table 2.7 Metric to Imperial mass
Metric
US or Imperial
1 g = 1000 mg
0.0353 oz
1 kg = 1000 g
2.2046 lb
1 tonne, t, = 1000 kg
1.1023 short ton
1 tonne, t, = 1000 kg
0.9842 long ton
The British ton is the long ton, which is 2240
pounds, and the U.S. ton is the short ton which is
2000 pounds.
Calculate the number of ounces in a
mass of 1346 g, correct to 2 decimal places.
1346 g = 1346 x 0.0353 oz = 47.51 oz from Table 2.7
Calculate the mass, in pounds, in a
210.4 kg mass, correct to 4 significant figures.
210.4 kg = 210.4 x 2.2046 lb = 463.8 lb from Table 2.7
Calculate the number of short tons
in 5000 kg, correct to 2 decimal places.
5000 kg = 5 t = 5 x 1.1023 short tons = 5.51 short tons
from Table 2.7
Further revisionary mathematics 27
Table 2.8 Imperial to Metric mass
US or Imperial
Metric
1 oz = 437.5 grain
28.35 g
1 lb = 16 oz
0.4536 kg
1 stone = 14 lb
6.3503 kg
1 hundredweight, cwt =
1121b 50.802 kg
1 short ton
0.9072 tonne
1 long ton
1.0160 tonne
Calculate the number of grams in
5.63 oz, correct to 4 significant figures.
5.63 oz = 5.63 x 28.35 g
= 159.6 g from Table 2.8
Calculate the number of kilograms
in 75 oz, correct to 3 decimal places.
75 oz = — lb = — x 0.4536 kg = 2.126 kg from
16
16
Table 2.8
Convert 3.25 cwt into (a) pounds
(b) kilograms.
(a) 3.25 cwt = 3.25 x 112 lb = 364 lb from Table 2.8
(b) 3.25 cwt = 3.25 x 50.802 kg - 165.1 kg from
Table 2.8
Temperature
To convert from Celsius to Fahrenheit, first multiply by
9/5, then add 32.
To convert from Fahrenheit to Celsius, first subtract 32,
then multiply by 5/9.
Convert 35°C to degrees Fahrenheit.
F = —C + 32 hence 35°C = -(35) + 32 = 63 + 32 = 95°F
5 5
Convert 113°F to degrees Celsius.
C= j(F - 32) hence 113°F = -(113-32) = —(81)
5 9 9
= 45°C
A summary of metric to imperial conversions, and vice
versa, is shown on page 328.
Now try the following Practice Exercise
Practice Exercise 12 Metric/imperial
conversions
In the following Problems, use the metric/impe¬
rial conversions in Tables 2.1 to 2.8
1. Calculate the number of inches in 476 mm,
correct to 2 decimal places [18.74 in]
2. Calculate the number of inches in 209 cm,
correct to 4 significant figures [82.28 in]
3. Calculate the number of yards in 34.7 m,
correct to 2 decimal places [37.95 yd]
4. Calculate the number of miles in 29.55 km,
correct to 2 decimal places
[18.36 miles]
5. Calculate the number of centimetres in
16.4 inches, correct to 2 decimal places
[41.66 cm]
6 . Calculate the number of metres in 78
inches, correct to 2 decimal places
[1.98 m]
7. Calculate the number of metres in 15.7
yards, correct to 2 decimal places
[14.36 m]
8 . Calculate the number of kilometres in 3.67
miles, correct to 2 decimal places
[5.91 km]
9. Calculate the number of (a) yards (b) kilo¬
metres in 11.23 nautical miles
[(a) 22,745 yd (b) 20.81 km]
10. Calculate the number of square inches in
62.5 cm 2 , correct to 4 significant figures
[9.688 in 2 ]
11. Calculate the number of square yards in
15.2 m 2 , correct to 2 decimal places
[18.18 yd 2 ]
12. Calculate the number of acres in 12.5
hectares, correct to 2 decimal places
[30.89 acres]
13. Calculate the number of square miles in
56.7 km 2 , correct to 2 decimal places
[21.89 mile 2 ]
Part One
Part One
28 Mechanical Engineering Principles
14. Calculate the number of square centi¬
metres in 6.37 in 2 , correct to the nearest
square centimetre [41 cm 2 ]
15. Calculate the number of square metres in
308.6 ft 2 , correct to 2 decimal places
[28.67 m 2 ]
16. Calculate the number of square metres
in 2.5 acres, correct to the nearest square
metre [10117 m 2 ]
17. Calculate the number of square kilometres
in 21.3 mile 2 , correct to 2 decimal places
[55.17 km 2 ]
18. Calculate the number of cubic inches in
200.7 cm 3 , correct to 2 decimal places
[12.24 in 3 ]
19. Calculate the number of cubic feet in
214.5 dm 3 , correct to 3 decimal places
[7.572 ft 3 ]
20. Calculate the number of cubic yards in
13.45 m 3 , correct to 4 significant figures
[17.59 yd 3 ]
21. Calculate the number of US fluid pints in
15 litres, correct to 1 decimal place
[31.7 fluid pints]
22. Calculate the number of cubic centimetres
in 2.15 in 3 , correct to 2 decimal places
[35.23 cm 3 ]
23. Calculate the number of cubic metres in
175 ft 3 , correct to 4 significant figures
[4.956 m 3 ]
24. Calculate the number of litres in 7.75 US
pints, correct to 3 decimal places
[3.667 litres]
25. Calculate the number of litres in 12.5 US
gallons, correct to 2 decimal places
[47.32 litres]
26. Calculate the number of ounces in 980 g,
correct to 2 decimal places [34.59 oz]
27. Calculate the mass, in pounds, in 55 kg,
correct to 4 significant figures [121.3 lb]
28. Calculate the number of short tons in
4000 kg, correct to 3 decimal places
[4.409 short tons]
29. Calculate the number of grams in 7.78 oz,
correct to 4 significant figures [220.6 g]
30. Calculate the number of kilograms in
57.5 oz, correct to 3 decimal places
[1.630 kg]
31. Convert 2.5 cwt into (a) pounds (b) kilograms
[(a) 280 lb (b) 127.2 kg]
32. Convert 55°C to degrees Fahrenheit [131°F]
33. Convert 167°F to degrees Celsius [75 °C]
2.3 Straight line graphs
A graph is a visual representation of information, show¬
ing how one quantity varies with another related quantity.
The most common method of showing a relationship
between two sets of data is to use a pair of reference
axes - these are two lines drawn at right angles to each
other, (often called Cartesian or rectangular axes), as
shown in Figure 2.1.
Figure 2.1
The horizontal axis is labelled the x-axis, and the verti¬
cal axis is labelled the y-axis.
The point where x is 0 andy is 0 is called the origin, x
values have scales that are positive to the right of the
origin and negative to the left.
y values have scales that are positive up from the origin
and negative down from the origin.
Co-ordinates are written with brackets and a comma
in between two numbers.
For example, point A is shown with co-ordinates (3, 2)
and is located by starting at the origin and moving 3
units in the positive x direction (i.e. to the right) and
then 2 units in the positive^ direction (i.e. up).
Further revisionary mathematics 29
When co-ordinates are stated, the first number is al¬
ways the x value, and the second number is always the
y value.
Also in Figure 2.1, point B has co-ordinates (-4, 3) and
point Chas co-ordinates (-3, -2)
The following table gives the force F newtons which,
when applied to a lifting machine, overcomes a corre¬
sponding load of L newtons.
F (newtons) 19 35 50 93 125 147
L (newtons) 40 120 230 410 540 680
1. Plot L horizontally and F vertically.
2. Scales are normally chosen such that the graph oc¬
cupies as much space as possible on the graph pa¬
per. So in this case, the following scales are chosen:
Horizontal axis (i.e. L): 1 cm = 50 N
Vertical axis (i.e. F): 1 cm = 10 N
3. Draw the axes and label them L (newtons) for the
horizontal axis and F (newtons) for the vertical
axis.
4. Label the origin as 0.
5. Write on the horizontal scaling at 100, 200, 300,
and so on, every 2 cm.
6 . Write on the vertical scaling at 10, 20, 30, and so
on, every 1 cm.
7. Plot on the graph the co-ordinates (40, 19),
(120, 35), (230, 50), (410, 93), (540, 125) and
(680, 147) marking each with a cross or a dot.
8 . Using a ruler, draw the best straight line through
the points. You will notice that not all of the points
lie exactly on a straight line. This is quite normal
with experimental values. In a practical situation
it would be surprising if all of the points lay ex¬
actly on a straight line.
9. Extend the straight line at each end.
10. From the graph, determine the force applied when
the load is 325 N. It should be close to 75 N. This
process of finding an equivalent value within the
given data is called interpolation.
Similarly, determine the load that a force of 45 N
will overcome. It should be close to 170 N.
11. From the graph, determine the force needed to over¬
come a 750 N load. It should be close to 161 N. This
process of finding an equivalent value outside the
given data is called extrapolation. To extrapolate
we need to have extended the straight line drawn.
Similarly, determine the force applied when the
load is zero. It should be close to 11 N.
Where the straight line crosses the vertical axis is
called the vertical-axis intercept. So in this case,
the vertical-axis intercept = 11 N at co-ordinates
( 0 , 11 )
The graph you have drawn should look something like
Figure 2.2 shown below.
L (newtons)
Figure 2.2
In another example, let the relationship between two
variables x and y be y = 3x + 2
Whenx = 0, y = 3 x() + 2 = 0 + 2 = 2
When x = 1, y = 3 xl+2=3+2=5
When x = 2,y = 3 x2 + 2 = 6 + 2 = 8, and so on.
The co-ordinates (0, 2), (1, 5) and (2, 8) have been
produced and are plotted, with others, as shown in
Figure 2.3.
Figure 2.3
Part One
Part One
30 Mechanical Engineering Principles
When the points are joined together a straight line
graph results, i.e. y = 3x + 2 is a straight line graph.
Now try the following Practice Exercise
Practice Exercise 13 Straight line graphs
1. Corresponding values obtained experimen¬
tally for two quantities are:
x -5
-3
-1
0
2
4
y -13
-9
-5
-3
1
5
Plot a graph of y (vertically) against x (hori¬
zontally) to scales of 2 cm = 1 for the hori¬
zontal x-axis and 1 cm = 1 for the vertical
y-axis. (This graph will need the whole of
the graph paper with the origin somewhere
in the centre of the paper).
From the graph find:
(a) the value of y when x = 1
(b) the value of y when x = -2.5
(c) the value of x when y = -6
(d) the value of x when y = 5
[(a) -1 (b) -8 (c) -1.5 (d) 4]
2. Corresponding values obtained experimen¬
tally for two quantities are:
X
-2.0
-0.5
0
1.0
2.5
3.0
5.0
y
-13.0
-5.5
-3.0
2.0
9.5
12.0
22.0
Use a horizontal scale forx of 1 cm = — unit
2
and a vertical scale fory of 1 cm = 2 units and
draw a graph of x against y. Label the graph
and each of its axes. By interpolation, find
from the graph the value ofy when x is 3.5
[14.5]
3. Draw a graph of y - 3x + 5 = 0 over a range
of x = -3 to x = 4. Hence determine (a) the
value of y when x = 1.3 and (b) the value of
x when y = -9.2 [(a) -1.1 (b) -1.4]
4. The speed n rev/min of a motor changes
when the voltage V across the armature is
varied.
The results are shown in the following table:
n (rev/min) 560 720 900 1010 1240 1410
V (volts) 80 100 120 140 160 180
It is suspected that one of the readings taken
of the speed is inaccurate. Plot a graph of
speed (horizontally) against voltage (verti¬
cally) and find this value. Find also (a) the
speed at a voltage of 132 V, and (b) the volt¬
age at a speed of 1300 rev/min.
[1010 rev/min should be 1070 rev/min;
(a) 1000 rev/min (b) 167 V]
2.4 Gradients, intercepts and
equation of a graph
Gradient
The gradient or slope of a straight line is the ratio
of the change in the value of y to the change in the
value of x between any two points on the line. If, as x
increases (—>), y also increases (T), then the gradient
is positive.
In Figure 2.4(a), a straight line graph y = 2x + 1 is
shown. To find the gradient of this straight line, choose
two points on the straight line graph, such as A and C.
y = 2x + 1
y 1
k
w
3
2
ll
UJ 1
1
i
1 1 b
0
1
2 3 x
(c)
Figure 2.4
Then construct a right angled triangle, such as ABC,
where BC is vertical and AB is horizontal.
Further revisionary mathematics 31
Then,
gradient of AC
change in y CB
change in x BA
_ 7-3
~ 3-1
2
In Figure 2.4(b), a straight line graph y = —3x + 2 is
shown. To find the gradient of this straight line, choose
two points on the straight line graph, such as D and F.
Then construct a right angled triangle, such as DEF,
where EF is vertical and DE is horizontal.
Then,
.. . f nr change in y FE
gradient of DF= - 5 — ; —— =-
change in x ED
_ 1 1 — 2 _ 9 _
-3-0 -3
Figure 2.4(c) shows a straight line graph y = 3. Since
the straight line is horizontal the gradient is zero.
y-axis intercept
Figure 2.5
(a) A right angled triangle ABC is constructed on the
graph as shown in Figure 2.6.
The value of y when x = 0 is called the y-axis inter¬
cept. In Figure 2.4(a) the y-axis intercept is 1 and in
Figure 2.4(b) the y-axis intercept is 2.
Gradient =
AC
CB
23-8
4-1
Equation of a straight line graph
The general equation of a straight line graph is:
y = mx + c
where m is the gradient or slope, and c is the y-axis
intercept
Thus, as we have found in Figure 2.4(a),y = 2x + 1 rep¬
resents a straight line of gradient 2 and y-axis intercept
1. So, given an equation;; = 2x + 1, we are able to state,
on sight, that the gradient = 2 and the y-axis intercept
= 1, without the need for any analysis.
Similarly, in Figure 2.4(b), y = —3x + 2 represents a
straight line of gradient -3 and y-axis intercept 2.
In Figure 2.4(c), y = 3 may be re-written as y = Ox + 3
and therefore represents a straight line of gradient 0
and y-axis intercept 3.
Here are some worked problems to help understanding
of gradients, intercepts and equation of a graph.
Determine for the straight line
shown in Figure 2.5:
(a) the gradient and (b) the equation of the graph
Figure 2.6
(b) The y-axis intercept at x = 0 is seen to be aty = 3.
y = mx + c is a straight line graph where
m = gradient and c = y-axis intercept.
From above, m — 5 and c — 3
Hence, equation of graph is: y = 5x + 3
Part One
Part One
32 Mechanical Engineering Principles
Determine the equation of the
straight line shown in Figure 2.7.
The triangle DEF is shown constructed in Figure 2.7.
DF 3-(-3) 6
Gradient of DE = -=-= — = -2
FE -1-2 -3
and the y-axis intercept = 1
Hence, the equation of the straight line is:
y = mx + c i.e. y = —2x + 1
Now try the following Practice Exercise
(c) y = - 3x - 4 (d) y = 4
[(a) 4,-2 (b) - 1, 0 (c) - 3, - 4 (d) 0, 4]
3. Draw on the same axes the graphs of
y — 3x - 5 and 3y + 2x = 7. Find the co-ordi¬
nates of the point of intersection.
[(2, 1)]
4. A piece of elastic is tied to a support so
that it hangs vertically, and a pan, on which
weights can be placed, is attached to the free
end. The length of the elastic is measured as
various weights are added to the pan and the
results obtained are as follows:
Load, W (N) 5 10 15 20 25
Length,/(cm) 60 72 84 96 108
Plot a graph of load (horizontally) against
length (vertically) and determine:
(a) the value of length when the load is 17 N,
(b) the value of load when the length is 74 cm,
(c) its gradient, and (d) the equation of the
graph.
[(a) 89 cm (b) 11 N (c) 2.4
(d) / = 2AW + 48]
5. The following table gives the effort P to lift
a load W with a small lifting machine:
Practice Exercise 14 Gradients, intercepts
and equation of a
graph
1. The equation of a line is 4y = 2x + 5. A table
of corresponding values is produced and is
shown below. Complete the table and plot a
graph ofy against x. Find the gradient of the
graph.
X -4
-3 -2
-1 0 1
2 3 4
y
-0.25
1.25
1.25
[Missing values: - 0.75, 0.25, 0.75, 1.75,
2.25, 2.75 Gradient = — ]
2
2. Determine the gradient and intercept on the
y-axis for each of the following equations:
(a) y = 4x - 2 (b) y = - x
W (N)
10
20
30
40
50
60
P (N)
5.1
6.4
8.1
9.6
10.9
12.4
Plot W horizontally against P vertically and
show that the values lie approximately on
a straight line. Determine the probable re¬
lationship connecting P and W in the form
P = aW + b. [P = 0A5W+ 3.5]
2.5 Practical straight line graphs
When a set of co-ordinate values are given or are
obtained experimentally and it is believed that they
follow a law of the form y = mx + c, then if a straight
line can be drawn reasonably close to most of the co¬
ordinate values when plotted, this verifies that a law of
the form y = mx + c exists. From the graph, constants
Further revisionary mathematics 33
m (i.e. gradient) and c (i.e. y-axis intercept) can be
determined.
Here is a worked practical problem.
In an experiment demonstrating
Hooke’s law, the strain in an aluminium wire was
measured for various stresses. The results were:
Stress N/mm 2
4.9
8.7
15.0
Strain
0.00007
0.00013
0.00021
Stress N/mm 2
18.4
24.2
27.3
Strain
0.00027
0.00034
0.00039
Plot a graph of stress (vertically) against strain
(horizontally). Find:
(a) Young’s Modulus of Elasticity for
aluminium which is given by the gradient
of the graph,
(b) the value of the strain at a stress of 20 N/mm 2 ,
(c) the value of the stress when the strain is
0 . 00020 .
The co-ordinates (0.00007, 4.9), (0.00013, 8.7), and
so on, are plotted as shown in Figure 2.8. The graph
produced is the best straight line which can be drawn
corresponding to these points. (With experimental re¬
sults it is unlikely that all the points will lie exactly
on a straight line.) The graph, and each of its axes, are
labelled. Since the straight line passes through the ori¬
gin, then stress is directly proportional to strain for the
given range of values.
The gradient of the straight line AC is given by
AB
5C
28-7
0.00040-0.00010
21 _ _ 1 _ _
3x 1(T 4 ltT*
21
0.00030
7 x 10 4
= 70000 N/mm 2
Thus, Young’s Modulus of Elasticity for aluminium
is 70000 N/mm 2 .
Since 1 m 2 = 10 6 mm 2 , 70000 N/mm 2 is equivalent to
70000 x 10 6 N/m 2 , i.e. 70 x 10 9 N/m 2 (or pascals).
From Figure 2.8:
(b) The value of the strain at a stress of 20 N/mm 2 is
0.000285, and
(c) The value of the stress when the strain is 0.00020
is 14 N/mm 2
N.B. 1 N/mm 2 = 1 MN/m 2 = 1 MPa
Now try the following Practice Exercise
Practice Exercise 15 Practical problems
involving straight line
graphs
1. The following table gives the force F new¬
tons which, when applied to a lifting ma¬
chine, overcomes a corresponding load of L
newtons.
Force F (newtons) 25 47 64 120 149 187
Load L (newtons) 50 140 210 430 550 700
Choose suitable scales and plot a graph of F
(vertically) against L (horizontally). Draw the
best straight line through the points. Deter¬
mine from the graph (a) the gradient, (b) the
F-axis intercept, (c) the equation of the graph,
(d) the force applied when the load is 310 N,
and (e) the load that a force of 160 N will
overcome, (f) If the graph were to continue in
the same manner, what value of force will be
needed to overcome a 800 N load?
[(a) 0.25 (b) 12 (c) F = 0.25L+12
(d) 89.5 N (e) 592 N (f) 212 N]
2. The following table gives the results of tests
carried out to determine the breaking stress
a of rolled copper at various temperatures, t :
Stress o (N/cm 2 ) 8.51 8.07 7.80 7.47 7.23 6.78
Temperature 75 220 310 420 500 650
f(°C)
Figure 2.8
Part One
Part One
34 Mechanical Engineering Principles
Plot a graph of stress (vertically) against
temperature (horizontally). Draw the best
straight line through the plotted co-ordinates.
Determine the slope of the graph and the
vertical axis intercept.
[-0.003, 8.73 N/cm 2 ]
3. The velocity v of a body after varying time
intervals t was measured as follows:
t (seconds) 2 5 8 11 15 18
v (m/s) 16.9 19.0 21.1 23.2 26.0 28.1
Plot v vertically and t horizontally and draw
a graph of velocity against time. Determine
from the graph (a) the velocity after 10 s,
(b) the time at 20 m/s and (c) the equation
of the graph.
[(a) 22.5 m/s (b) 6.5 s (c) v - 0.7 1 + 15.5]
4. An experiment with a set of pulley blocks
gave the following results:
Effort, E
(newtons) 9 11 13.6 17.4 20.8 23.6
Load, L
(newtons) 15 25 38 57 74 88
Plot a graph of effort (vertically ) against load
(horizontally) and determine (a) the gradi¬
ent, (b) the vertical axis intercept, (c) the law
of the graph, (d) the effort when the load is
30 N and (e) the load when the effort is 19 N.
[(a) I or 0.2 (b) 6 (c) E = 0.2L + 6
5 (d) 12 N (e) 65 N]
2.6 Introduction to calculus
Calculus is a branch of mathematics involving or lead¬
ing to calculations dealing with continuously varying
functions - such as velocity and acceleration, rates of
change and maximum and minimum values of curves.
Calculus has widespread applications in science and
engineering and is used to solve complicated problems
for which algebra alone is insufficient.
Calculus is a subject that falls into two parts:
(i) differential calculus (or differentiation) and
(ii) integral calculus (or integration)
2.7 Basic differentiation revision
Differentiation of common functions
The standard derivatives are summarised in Table 2.9
and are true for all real values of x.
Table 2.9 Standard Differentials
y or fix)
dy or f'(x)
dx
ax 11
n —1
anx
sin ax
a cos ax
cos ax
—a sin ax
Q ax
a Q ax
In ax
1
X
In Table 2.9, — is the gradient, or slope, of 4 y ’ with
respect to V. ^
The differential coefficient of a sum or difference is
the sum or difference of the differential coefficients of
the separate terms.
Thus, if f[x) =p(x) + q(pc) - r(x), (where/ p , q and r are
functions), then f\x) =p' (x) + q (x) - r (x)
Differentiation of common functions is demonstrated
in the following worked problems.
Find the differential coefficients of:
(a)j=12 x 3 (b)jl
Ifv
(a)
: ax n then — = anx n 1
dx
Since y = 12x 3 , a = 12 and n
4y
dx
= (12)(3)x 3-1 = 36x 2
thus
(b) y
12 .
is rewritten in the standard ax 11 form as
x‘
y = 12x 3 and in the general rule a = 12 and
n = -3
Thus ^ =(12)(-3)x- 3 -'=-36x^ = -^
dx x
Differentiate: (a) y = 6 (b) y = 6x
(a) y — 6 may be written as y = 6x°, i.e. in the general
rule a = 6 and n = 0.
Hence = (6)(0)x 0-1 = 0
dx
Further revisionary mathematics 35
In general, the differential coefficient of a constant
is always zero.
(b) Since y = 6 x, in the general rule a — 6 and n = 1
Hence = ( 6 )(l)x 1_1 = 6 x° = 6
dx
In general, the differential coefficient of kx,
where k is a constant, is always k.
Problem 42.
Find the derivatives of:
(a)j = 3 77
(b + 3 rr
Vx;
(a) y — 3 V* is rewritten in the standard differential
form asy = 3x 1/2
In the general rule, a = 3 and n — —
/ \
dy
Thus — = (3)
dx
1
V 2 .
— 1
x
3
— x
2
2
2x
1/2
2^[x
(b) y
4/3
= 5x 4/3 in the standard differen-
7+ x
tial form.
4
In the general rule, a = 5 and n = - —
/ \ 3
Thus
4y
dx
(5)
-20
v
4
3
x
(-4/3)—1 =
7
-20
3x
7/3
37T
-20
x
-7/3
Find the differential coefficients of:
(a) y = 3 sin 4x (b) /(/) = 2 cos 3/ with respect to
the variable
(a) Wheny = 3 sin 4x, a = 4, so that
— = (3)(4 cos 4x) = 12 cos 4x from Table 2.9
dx
(b) When/(/) = 2 cos 3/, a = 3, so that
f\t) = (2) (-3 sin 3/) = -6 sin 3 1
Determine the derivatives of:
(a) y = 3e5x (b)/(0) = \ (c) y = 6 In 2x
(a) Wheny = 3e 5v , a = 5, so that
(b) /( 0 )
^ = (3)(5)e Sjc =
dr 2
15e 5A ‘ from Table 2.9
~2e 3e , a = - 3, so that
.30
-30-
(c)
f\ 0) = (2) (-3)e -30 = - 6 e
dv
When y = 6 In 2x then — = 6
dx
-6
A
1
V X 7
Find the gradient of the curve
y = 3x 4 - 2x 2 + 5x - 2 at the points (0, - 2) and (1,4).
The gradient of a curve at a given point is given by
the corresponding value of the derivative. Thus, since
y= 3x 4 - 2x 2 + 5x - 2 then the gradient = — = 12x 3 - 4x + 5
dx
At the point (0, - 2), x = 0.
Thus the gradient = 12(0 ) 3 - 4(0) + 5 = 5
At the point (1, 4), x = 1.
Thus the gradient = 12(1 ) 3 - 4(1) + 5 = 13
Now try the following Practice Exercise
Practice Exercise 16 Further problems on
differentiating com¬
mon functions
In Problems 1 to 6 find the differential coeffi¬
cients of the given functions with respect to the
variable.
1 .
(a)
5x 5 (b) -
[(a)
25x 4 (b)
-T]
X
[(a) 8 3
x 2
2 .
(a)
~ 4 (b) 6
(b) 0 ]
z
X
X
' i
vx
3.
(a)
2 x (b) 2 77
[(a) 2 (b)
4.
(a)
37x 5 (b)
4
i —
[(a) 5 a
Jx 2 (b)--
?-]
Vx
\lx 3
5.
(a)
2 sin 3x (b)
- 4 cos 2x
[(a) 6 cos 3x
(b) 8 sin 2 x]
6 . (a) 2e 6 * (b) 4 In 9x [(a) 12e 6 * (b) - ]
x
7. Find the gradient of the curve y = 2 / 4 + 3 / 3 -
t + 4 at the points (0, 4) and (1,8)
[- 1 , 16]
8 . (a) Differentiate
y= +2ln20-2(cos 56+ 3 sin20)-
0
39
7 /jj-
(b) Evaluate when 6 = — , correct to 4
d0 2
significant figure.
[(a) -4- + — + 10 sin 56-
6 3 6
12 cos 2 0+ \ (b) 22.30]
x
Part One
Part One
36 Mechanical Engineering Principles
2.8 Revision of integration
The general solution of integrals of the
form ax n
The general solution of integrals of the form J ax' 1 dx,
where ‘a’ and ‘n’ are constants and n& — 1 , is given by:
f n . aX " +l
I ax dx = - + c
J n +1
Note that J ax" dx actually represents the area under
the curve y = ax n and c is an arbitrary constant.
Integrals of this type are called indefinite integrals.
(Section 2.9 below explains definite integrals where
limits are applied to the integral, for example, ax" dx)
" a
Using the above rule gives:
(i) For j* 3x 4 dx, a = 3 and n = 4, so that
J 3x 4 dx =
3x
4+1
+ C = —X* + c
4 + 1 5
r 4 3 4
(ii) For J—tdt,a=—,n = 3 and t = x, so that
+ c = -t 4 + c
9
'xf 1
t-i
-o
r<l
'xf |
(?*]
4
+ c = —
j 9 9
V + 1 J
9
l 4 J
(iii) For f — dx = f 2x ' 2 dx, a = 2 and n = -2, so
J x" J
2 x 1
r x “ 2+1 2 x
that / 2 x 2 dx = ( 2 )-+ c = -+ c
J -2 + 1 -1
-1
2
-+ c, and
x
(iv) For J Vx d x = Jx 2 dx, a = 1 and n = —,
so
i
that J" x^ dx = ( 1 )
x
i+i
2
X
+ c =
+ 1
3
2
+ c
= —\[x* + c
3
Each of these results may be checked by differentiation.
(a) The integral of a constant k is kx + c.
For example, J 8 dx = 8 x + c and J 5 At = 5t + c
(b) When a sum of several terms is integrated the result
is the sum of the integrals of the separate terms. For
example,
J"^3x + 2x 2 — 5jdx = 3xdx+ J 2x 2 dx- J* 5 dx
= +1 + - 5x + c
Standard integrals
Since integration is the reverse process of differentia¬
tion the standard integrals listed in Table 2.10 may be
deduced and readily checked by differentiation.
Table 2.10 Standard Integrals
y
Jydx
1.
o
n+l
n qx
/ ax" - + c (except when n = - 1 )
' n +1
2 - J
f cos ax dx — sin ax + c
3.
o
r • l
/ sin ax dx - — cos ax + c
' a
4.
fj
r e at dx -e ax + c
1 a
— dx lnx + c
X
Problem Determine: J lx 2 dx
The standard integral, J ax d x
When a — 1 and n = 2,
7x 2+1
ax
«+i
n + l
+ c
J 7x 2 dx
7 ,
lx 3
+ c = - + c or ljc j + c
2 + 1 3 3
Proble Determine: J2 ^ 3 d^
When a = 2 and n — 3,
/
^ s i 2^ 3+1 It" 1 4
2 1 dt — - + c — — + c — — /^ + c
3+1 4 2
Note that each of the results in worked examples 46
and 47 may be checked by differentiating them.
Problem 48. Determine: J 8 dx
1 8 dx is the same as J 8 x° dx and, using the general
rule when a — 8 and n = 0 gives:
8 x 0+I
J 8 x° dx
0+1
+ c = 8 x + c
In general, if A: is a constant then J k dx = kx + c
Further revisionary mathematics 37
Problem 49. Determine: J*2xdx
When a = 2 and n = 1, then
^ i+i
lx
J 2 xdx
Determine: f dx
J 5x
f 2xdx = J 2 x' dx
2 x 2
+ c = -+ C = X +c
1+1 2
Note that when an integral contains more than one term
there is no need to have an arbitrary constant for each;
just a single constant c at the end is sufficient.
Determine: J* -^dx
x 2
J -^ r dx = J 5x 2 dx
x
n = - 2 gives:
r s>r~ 2+l
f 5x~ 2 dx = —— +
J - 2+1
5x
-i
-1
5
x
Problem 51 Determine: J 3^fxdx
/ in —
V a= a n - see Chapter 1, page 12.
J 3^fxdx = J 3x 2 dx = —
i+i
3x 2
+ 1
3
3x 2
+ C=— +c
2
= 2x 2 + c = 2 Vx 3 " + c
Problem Determine: J* 4cos3xdx
From 2 of Table 2.10,
J 4cos3xdx = (4)
V
V 3 7
sin 3x + c = — sin 3 x + c
3
Problem 53. Determine: J* 5 sin 2 OdO
From 3 of Table 2.10,
J 5sin2 OdO = ( 5 )
/
v
1 I 5
— cos 20+ c = -cos 2 6+ c
2 j 2
From 4 of Table 2.10,
/ \
r II
J 5e 3A dx = (5)
V 3 7
q3x _j_ c — _ e 3x c
3
From 5 of Table 2.10,
/+=/
3
5
/ \
1
dx = - In x + c
5
Now try the following Practice Exercise
Practice Exercise 17 Standard integrals
Determine the following integrals:
1 a = 5 and
1 . (a) _
f 4dx
(b) J7xdx
[(a) 4x + c (b) —x 2 + c ]
2
1 + c
2. (a) _
^ 5x 3 dx
(b) JVdf
+ c
[(a) —x 4 + c (b) -/ 8 +c]
4 8
3. (a) _
^ |x 2 dx
(b) f — x 3 dx
J 6
appreciate
[(a) — x 3 +c (b) — x 4 + c]
15 24
4. (a) /< 2x 4 — 3x j dx (b) J* (2 — 3 t 3 j d^
[(a) — x 5 -— x 2 + c (b) 2 t-—t 4 +c\
5 2 4
(a) J
3x'
dx
<b)/
4x‘
dx
[(a) -1 + c (b) -Ty + c]
3x 4x
6. (a) 2 J Vx 3 " dx (b) dx
[( a ) —Vx^ + C (b) —yfx^ + c]
5 9
7. (a) J 3cos2xdx (b) J"7sin3 OdO
3 7
[(a) —sin2x + c (b) —cos30 + c]
— 3
8 . (a) J 3sin — xdx (b) J 6 cos-
-xdx
3
[(a) -6cos—x + c (b) 18sin-x + c ]
2 3
Part One
Part One
38 Mechanical Engineering Principles
9.
(a) f - e 2v dx
J 4
<b, /
dx
3x
[(a) h 2x +c (b) hnx + c]
2.9 Definite integrals
Integrals containing an arbitrary constant c in their re¬
sults are called indefinite integrals since their precise
value cannot be determined without further information.
Definite integrals are those in which limits are applied.
If an expression is written as [x] , ‘Z?’ is called the
upper limit and 6 a 9 the lower limit.
The operation of applying the limits is defined as:
H!
For example, the increase in the value of the integral
x 2 as x increases from 1 to 3 is written as f x 2 dx,
J i
where the lower limit is 1, and the upper limit is 3.
Applying the limits gives:
r 3 7
3
X
3
f 3 3
\
(V ")
/ x dx =
— + c
—
hC
J i
3
3
3
1
V
/
V 7
1
—hC
3
2
8 3
= (9 + c)-
V“ /
Note that the V term always cancels out when limits are
applied and it need not be shown with definite integrals.
r 2 n
~3x 2 "
J 3x dx =
2
= {!(2) 2 } - { XI) 2 } = 6 - A
r» 2 n 2
/ x(3 + 2x)dx = / (3x + 2x 2 )dx
Jo Jo
3x 2x‘
- +—
~i2
Jo
3(2)' ' 2(2)
-{o+o)
16 1
= 6+ — =11- or 11.33
3 3
f >r/2
Evaluate: 3sin2xdx
J o
r> K/2
/ 3sin2xdx =
(3)
cos2x
J 0
, 2,
-i k /2
-17T/2
—cos2x
2
{- - cos 2
Jo
J,
} - {--COS 2(0)}
3 3
= {- -cos 71 } - {--cos 0}
= {- 2 H )}-{- 2 ( 1 )}
3 3
— — + — — 3
2 2
Evaluate: 4 cos 3^
f 4cos3^d/ =
(4)
/ \
1
sin 3 1
2
4
— sin 3 1
) i
,7
1
_ 3
-i 2
J 1
z z z
1
4 . ^
4 . „
<
—sin 6
* •<
—sin 3
1
3
3
Evaluate: J (4-x 9 )dx
f 3 (4 —x 2 )dx=
n3
4x —
x
-2
3
Note that limits of trigonometric functions are al¬
ways expressed in radians - thus, for example, sin 6
means the sine of 6 radians = -0.279415.... where n
radians - 180°
Hence.
(3)\
= {4(3) - — } —{4(-2)
= { 12-9}-{- 8 - y }
= {3}-{-5^}=8^
(- 2 ) :
r 2
4
J 4cos3^d/ = <
-(-0.279415..)
3 '
}
-(0.141120..)
3
= (-0.37255) - (0.18816)
= -0.5607
Further revisionary mathematics 39
Z,
Evaluate: J 4e ' dx correct to 4
significant figures.
f 4e 2 ' dx = —e 2x =2 e 2x =2[e 4 -e 2 l
J i 9 1
^ i
= 2[54.5982 - 7.3891]
- 94.42
Evaluate:
significant figures.
3
— d u correct to 4
4 u
— [In 4 - In 1]
4
3
— [1.3863 - 0] = 1.040
4
Now try the following Practice Exercise
Practice Exercise 18 Definite integrals
In Problems 1 to 7, evaluate the definite integrals
(where necessary, correct to 4 significant figures).
1 .
(a)
J " xdx
(b)
jd x - ] ) dx
[(a) 1.5 (b) 0.5]
2 .
(a)
r* 4
J 5x 2 dx
(b)
I”-!,’d<
J -i 4
[(a) 105 (b) -0.5]
3.
(a)
f](3~x 2 )dx
(b)
J ^x 2 — 4x + 3j dx
[(a) 6 (b) -1.333]
4.
(a)
f 2 >/x dx
J 0
(b)
f\ dx
J 2 X 2
[(a)
10.67 (b) 0.1667]
5.
(a)
n x 3
/ — cos 0 d 0
do 2
(b)
n k!2
/ 4cos0d0
0
^ [(a) 0 (b) 4]
(a) f 2sin 20dO (b) f 3sin/ dt
J n !6 J 0
[(a) 1 (b) 4.248]
7.
<■) i:
3/ ,. _ r 3 2
3e dt
< b >i
dx
2 3x
[(a) 19.09 (b) 0.2703]
2.10 Simple vector analysis
Determinants
For a 2 by 2 determinant:
a b
c d
= ad- be
For example,
5 -3
4 2
For a 3 by 3 determinant:
(5)(2)-(-3)(4)= 10—12
= 22
a b c
d e f
g h i
a
e f
h i
-b
d f
g i
+ c
d e
g h
= a(e x i -f x h) - b(d x / -f x g) + c( d x h-e* g)
For example,
3 1 -4
-5 2 1
6 -3 -1
= 3
2 1
-5 1
-5 2
-3 -1
-1
6 -1
+ M)
6 -3
3(-2—3)- 1(5-6)-4(15- 12)
3(1) - 1(-1) - 4(3) = 3 + 1 — 12 = — 8
i, j, k unit vectors
A method of completely specifying the direction of
a vector in space relative to some reference point is
to use three unit vectors, i, j and k , mutually at right
angles to each other, as shown in Figure 2.9.
t z
—►
y
Figure 2.9
Part One
Part One
40 Mechanical Engineering Principles
The unit vectors are defined by: i = (l,0,0) y = (0,1,0)
* = ( 0 , 0 , 1 )
The scalar or dot product of two vectors
Let a = a { i + a 2 j + a^k and b = b { i + b 2 j + b 3 k
then a • b = (a { i + a 2 j + a 3 k) • ( b { i + b 2 j + b 2 k)
Multiplying out the brackets gives:
a • b = a { b { i • i + a { b 2 i • j + a { b 2 i • k + a 2 b { j • i
+ a 2 b 2 j • j + a 2 bij • k + a 2 b { k • i + a^b^k • j
+ a^b^k • k
However, it may be shown that
i • i =j • j = k • k = 1 and
i •/' = i*k=j*i=j*k = k*i = k*j = 0
Thus, the scalar or dot product,
a • b = a { b x + + ^ 3^3
For example, if a = 2i +j - k and b = i - 3 j + 2k
then a *b = (2)(1) + (1)(-3) + (- l)(2)=-3
A further example: Calculate the work done by a
force F = (-5 i + j + 7A) N when
its point of application moves
from point (-2 i - 6j + k) m to the
point (i -j + 10&) m.
Work done = F • d where d = ( i-j + 10 k) - (-2 i-6j + k)
= 3 1 + 5/ + 9k
Hence, work done = (-5/ +j+ Ik) • (3i + 5 j + 9k)
= -15 + 5 + 63 = 53 N m
The vector or cross product of two vectors
When a = a { i + a 2 j + a 3 k and b = b { i + b 2 j + b 3 k
then it may be shown that
i j k
the vector or cross product, a x b
a 2 a 3
\ b 2 b 3
For example, if a = i + 4 j - 2k and b = 2i-j + 3 k
then a x b =
i j k
1 4 -2
2-13
4 -2
-1 3
1 -2
2 3
+ k
1 4
2 -1
i(12 -2) -7(3 --4) + *(-1-8)
10/ - Ij - 9k
A further example: A force of (2/ —7 + Ar) newtons acts
on a line through point P having
co-ordinates (0, 3, 1) metres.
Determine the moment vector
about point Q having co-ordinates
(4, 0,-1) metres.
Position vector, r = (0/ + 3 j +k) - (4 i + 0 j - k)
= -4 i + 3 j + 2k
Moment, M = r x F where
= (3 + 2)i - (-4 - 4)j + (4 - 6)k
= (5/ + 87 - 2k) N m
Now try the following Practice Exercises
Practice Exercise 19 Simple vector analysis
1. Calculate the value of
2 -3
5 1
2. Evaluate
1 2
-5 -7
3.
Find the value of
2
-5
1
3
0
-4
-1
4
-2
[3]
[- 6 ]
Find the value of
-3 2 5
7 1 -2
4 0-1
[-19]
5. Find the scalar product a*b when
a = 3i + 2j - k and b = 2i + 5j + k [15]
6 . Find the scalar product p • q when
p = 2i - 3/ + 4k and q = i + 2j + 5k [16]
7. Given p = 2i- 3 j, q = 4j- k and
r = 1 + 2j - 3k, determine the quantities
(a) p • q (b ) p-r (c ) q-r
[(a)-12 (b) - 4 (c) 11]
8 . Calculate the work done by a force
F = (-2 1 + 3 j + 5k) N when its point of ap¬
plication moves from point (-1 - 5 j + 2k)
m to the point (4 1 - j + 8 k) m [32 N m]
Further revisionary mathematics 41
9. Given that p = 3i + 2k, q = i-2j + 3k and
r = —4 i + 3 j — k determine
(a) pxq (b) q x r
[(a) 4 1 - Ij - 6k (b) -li - 11/ - 5£]
10. A force of (4 i - 3 j + 2k) newtons acts on
a line through point P having co-ordinates
(2, 1, 3) metres. Determine the moment
vector about point Q having co-ordinates
(5, 0, -2) metres.
[M= (17/ + 26/ + 5&) N m]
Practice Exercise 20 Multiple-choice ques¬
tions on further revi¬
sionary mathematics
(Answers on page 335)
1. Differentiating y = 4x gives:
2 f. d y
(,)
W dr 3
(O fW
dx
(b) — = 20 x‘
dx
(d, f = 5,*
dx
Figure 2.10
6 . A graph of y against x, two engineering
quantities, produces a straight line.
A table of values is shown below:
X
2
-1
p
y
9
3
5
2. J (5 —3/ 2 )dMs equal to:
(a) 5-t 3 +c (b) -3; 3 + c
(c) -6t + c (d) 5 1 -f + c
3. A graph of resistance against voltage for an
electrical circuit is shown in Figure 2.10.
The equation relating resistance R and volt¬
age V is:
(a) R= 1.45 V + 40 (b) R = 0.8 V + 20
(c) R= 1.45 V+ 20 (d) R= 1.25 V + 20
4. The gradient of the curve y = -2x 3 + 3x + 5
atx = 2 is:
(a) -21 (b) 27
(c) -16 (d) -5
5. Thevalueof f (3sin2^ —4cos0)d9,correct
J o
to 4 significant figures, is:
(a) -0.06890 (b) -1.242
(c) -2.742 (d) -1.569
7.
The value of p is:
(a) ~ (b) -2 (c) 3 (d)0
dv .
dx
is equal to:
(b) 2 V? — 2 x + .
(d) 5Vx-2x
to:
(a) L. + c
18
(c) + c
9
(b) ^2 +c
3
(d) -t 3 +c
9
9. A water tank is in the shape of a rectangular
prism having length 1.5 m, breadth 60 cm
and height 300 mm. If 1 litre = 1000 cm 3 4 5 , the
capacity of the tank is:
(a) 27 litre (b) 2.7 litre
(c) 2700 litre (d) 270 litre
Part One
Part One
42 Mechanical Engineering Principles
10 .
n
11
12 .
13,
14.
15.
Evaluating f 3 3sin3xdx gives:
J e
(a) 1.503 (b) - 18 (c) 2 (d) 6
Here are four equations in x and y. When x is
plotted against y, in each case a straight line
results.
(i) y + 3 = 3x (ii) y + 3x = 3
,...v y 3 y 2
(ill)-= X (iv) — = x + —
2 2 3 3
Which of these equations are parallel to
each other?
(a) (i) and (ii) (b) (i) and (iv)
(c) (ii) and (iii) (d) (ii) and (iv)
Differentiating i = 3 sin 2t - 2 cos 3 1 with
respect to t gives:
(a) 3 cos 2t + 2 sin 3 1 (b) 6 (sin It- cos 3/)
3 2
(c) — cos 2t + — sin 3t (d) 6 (cos 2 t + sin3f)
J* (5 sin 3/ — 3 cos 5/) dt is equal to:
(a) -5 cos 3^ + 3 sin 5 1 + c
(b) 15(cos 3^ + sin 3 1) + c
5 3
(c) --cos3^--sin5/ + c
3 5
3 5
(d) -cos3^--sin5/ + c
5 3
Which of the straight lines shown in Figure
2.11 has the equation y + 4 = 2x?
(a) (iv) (b) (ii) (c) (iii) (d) (i)
Giveny = 3e x +21n3x, — is equal to:
dx
(a) 6 e x +
3x
(b) 3e x + —
x
16.
17.
(d) 3e x + —
3
(c) 6 e x + —
x
Given f(?) = 3t 4 - 2, f '(0 is equal to:
3
(a) 12/ 3 — 2
(c) 12 1 3 (d) 3t 5 -2
(b) — t - 2t + c
4
f
l vJ
dx is equal to:
(iii)
(iv)
Figure 2.11
(a)
8
2x
+ C
(b) x +
2x
(C ) X - \ + c (d) x —+ c
18. In an experiment demonstrating Hooke’s
law, the strain in a copper wire was measured
for various stresses. The results included
Stress (meg¬
apascals)
18.24
24.00
39.36
Strain
0.00019
0.00025
0.00041
When stress is plotted vertically against
strain horizontally a straight line graph
results. Young’s modulus of elasticity for
copper, which is given by the gradient of
the graph, is:
(a) 96000 Pa (b) 1.04 x 10" 11 Pa
(c) 96 Pa (d) 96 x 10 9 Pa
19. The gradient of the curve y = 4x 2 -7x + 3 at
the point ( 1 , 0 ) is
(a) 3 (b) 0 (c) 1 (d) -7
/ 2 ^
( 2e ' d£, correct to 4 signifi¬
cant figures, gives:
(a) 2300 (b) 255.6 (c) 766.7 (d) 282.3
For fully worked solutions to each of the problems in Practice Exercises 11 to 20 in this chapter,
go to the website:
www.routledge.com/cw/bird
Revision Test 2 Further revisionary mathematics
This Revision Test covers the material in Chapter 2. The marks for each question are shown in brackets at the end
of each question.
1. If 12 inches = 30.48 cm, find the number of mil¬
limetres in 23 inches. (3)
2. State the SI unit of:
(a) force (b) pressure (c) work (3)
3. State the quantity that has an SI unit of:
(a) kilograms (b) rad/s
(c) hertz (d) m 3 (4)
4. Express the following in engineering notation in
prefix form:
(a) 250,000 J (b) 0.00005 s
(c) 2 x 10 8 W (d) 750 x 1(T 8 F (4)
5. Rewrite:
(a) 0.0067 mAin pA
(b) 40 x 10 4 kNasMN (2)
6 . Determine the value of P in the following table of
values
x 0 14
y = ?>x- 5 -5 -2 P (2)
7. Corresponding values obtained experimentally
for two quantities are:
x -5-3-1024
y -17 -11 -5 -2 4 10
Plot a graph of (vertically) against x (horizontal¬
ly) to scales of 1 cm = 1 for the horizontal x- axis
and 1 cm = 2 for the vertical v-axis.
✓
From the graph find:
(a) the value of 32 when x = 3
(b) the value of y when x = - 4
(c) the value of x when y = 1
(d) the value of x when y = - 20 ( 8 )
9. The resistance R ohms of a copper winding is
measured at various temperatures t°C and the
results are as follows:
R(Q) 38 47 55 62 72
t(°C) 16 34 50 64 84
Plot a graph of R (vertically) against t (horizon¬
tally) and find from it:
(a) the temperature when the resistance is 50 Q.
(b) the resistance when the temperature is 72°C
(c) the gradient
(d) the equation of the graph. ( 10 )
10. Differentiate the following functions with respect
to x:
(a) y = 5x 2 - 4x + 9
(b) y = x 4 - 3x 2 — 2
(4)
3 . d y
11.
If y = — determine —
(2)
x dx
12.
Given f(f) = Vh , find f '(t)
(2)
13.
X
Calculate the gradient of the curve y = 3 cos —
n
3
at x = ~ correct to 3 decimal places.
(4)
14.
Find the gradient of the curve f(x) = lx 2
- 4x + 2
at the point (1,5)
(3)
15.
If y = 5 sin 3x - 2 cos 4x find —
(2)
dx
16. Determine the value of the differential coefficient
3
of y = 5 In 2x - “77 when x = 0.8 correct to 3
e" x
significant figures. (4)
8 . If graphs of y against x were to be plotted for In Problems 17 to 19, determine the indefinite integrals.
each of the following, state (i) the gradient, and
(ii) the 32 -axis intercept.
17. (a)
J ^x 2 + 4jdx
0» / ‘
u X
dx
(4)
(a) y = -5x + 3
(b) y = lx
2
f
\
1
-2
3x
/
(c) 2y + 4 = 5x
(d) 5x + 2 y = 6
( 8 )
18. (a)
j^dv (b)
I
0.5x
e +
<
dx
( 6 )
Part One
Part One
Further revisionary mathematics
19. (a) J(2 + d) 2 d0
(b) /
1 3
cos — x + — - e
2 x
2x
dx
( 6 )
Evaluate the integrals in problems 20 to 22, each,
where necessary, correct to 4 significant figures:
20
21
. (a) f'(t 2 -2t)dt
2
(b) /: (lx 3 — 3x ? + 2jdx
n n/3 3 ji/4 1
. (a) / 3sin2^d£ (b) / cos-x dx
j 71/4 3
( 6 )
(7)
22. (a) J* |Vx+2e v jdx
(b)
/,
/ \
r —
V r 7
dr
( 6 )
23. Evaluate (a)
5 -3
4 2
(b)
3
-5
6
1
2
-3
-4
1
-1
24. If a = 2i + 4 j - 5k and b = 3i - 2j + 6k deter¬
mine:
(i) a*b (ii) a x b (7)
25. Determine the work done by a force of F newtons
acting at a point A on a body, when A is displaced
to point B, the co-ordinates of A and B being
(2, 5, -3) and (1, -3, 0) metres respectively, and
when F=2i- 5j + 4k newtons. (5)
26. A force F=3i — 4 j + k newtons act on a line pass¬
ing through a point P. Determine moment M of
the force F about a point Q when P has co-ordi¬
nates (4, — 1, 5) metres and Q has co-ordinates
(4, 0, -3) metres. (6)
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 2,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
Pa rt Two
This page intentionally left blank
Chapter 3
The effects of forces
on materials
Why it is important to understand: The effects of forces on materials
A good knowledge of some of the constants used in the study of the properties of materials is vital in most
branches of engineering, especially in mechanical, manufacturing, aeronautical and civil and structural
engineering. For example, most steels look the same, but steels used for the pressure hull of a submarine
are about 5 times stronger than those used in the construction of a small building, and it is very important
for the professional and chartered engineer to know what steel to use for what construction; this is because
the cost of the high-tensile steel used to construct a submarine pressure hull is considerably higher than
the cost of the mild steel, or similar material, used to construct a small building. The engineer must not
only take into consideration the ability of the chosen material of construction to do the job, but also its
cost. Similar arguments lie in manufacturing engineering, where the engineer must be able to estimate the
ability of his/her machines to bend, cut or shape the artefact s/he is trying to produce, and at a competitive
price! This chapter provides explanations of the different terms that are used in determining the properties
of various materials. The importance of knowing about the effects of forces on materials is to aid the design
and construction of structures in an efficient and trustworthy manner.
At the end of this chapter you should be able to:
• define force and state its unit
• recognise a tensile force and state relevant practical examples
• recognise a compressive force and state relevant practical examples
• recognise a shear force and state relevant practical examples
• define stress and state its unit
p
• calculate stress o from o - —
A
• define strain
• calculate strain s from s = —
J-j
• define elasticity, plasticity, limit of proportionality and elastic limit
• state Hooke’s law
• define Young’s modulus of elasticity E and stiffness
• appreciate typical values for E
• calculate E from E - —
£
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Two
48 Mechanical Engineering Principles
• perform calculations using Hooke’s law
• plot a load/extension graph from given data
• define ductility, brittleness and malleability, with examples of each
• define rigidity or shear modulus
• understand thermal stresses and strains
• calculates stresses in compound bars
A force exerted on a body can cause a change in either
the shape or the motion of the body. The unit of force
is the newton*, N.
No solid body is perfectly rigid and when forces are
applied to it, changes in dimensions occur. Such changes
are not always perceptible to the human eye since they are
so small. For example, the span of a bridge will sag under
the weight of a vehicle and a spanner will bend slightly
when tightening a nut. It is important for engineers and
designers to appreciate the effects of forces on materials,
together with their mechanical properties.The three main
types of mechanical force that can act on a body are:
(i) tensile (ii) compressive and (iii) shear
*Sir Isaac Newton (25 December 1642-20 March 1727) was an
English polymath. His single greatest work, the ‘Philosophiae
Naturalis Principia Mathematica’ (‘Mathematical Principles of
Natural Philosophy’), showed how a universal force, gravity,
applied to all objects in all parts of the universe. To find out
more go to www.routledge.com/cw/bird
Tension is a force that tends to stretch a material, as
shown in Figure 3.1. For example,
(i) the rope or cable of a crane carrying a load is in
tension
<-
Force
—►
Force
Figure 3.1
(ii) rubber bands, when stretched, are in tension
(iii) when a nut is tightened, a bolt is under tension
A tensile force, i.e. one producing tension, increases
the length of the material on which it acts.
3.3 Compressive force
Compression is a force that tends to squeeze or crush a
material, as shown in Figure 3.2. For example,
- >
Force
Force
Figure 3.2
(i) a pillar supporting a bridge is in compression
(ii) the sole of a shoe is in compression
(iii) the jib of a crane is in compression
A compressive force, i.e. one producing compression,
will decrease the length of the material on which it acts.
3.4 Shear force
Shear is a force that tends to slide one face of the
material over an adjacent face. For example,
(i) a rivet holding two plates together is in shear if
a tensile force is applied between the plates - as
shown in Figure 3.3
The effects offerees on materials 49
Rivet
Force
-<—
W/
M///////M
Force
Figure 3.3
(ii) a guillotine cutting sheet metal, or garden shears,
each provide a shear force
(iii) a horizontal beam is subject to shear force
(iv) transmission joints on cars are subject to shear
forces
A shear force can cause a material to bend, slide or twist.
Figure 3.4(a) represents a crane and
Figure 3.4(b) a transmission joint. State the types
of forces acting, labelled A to F.
A rectangular bar having a cross-
sectional area of 75 mm 2 has a tensile force of
15 kN applied to it. Determine the stress in the bar.
Cross-sectional area A- 75 mm 2 = 75 x 10 6 m 2
and force F = 15kN= 15xl0 3 N
Stress in bar,
F 15 x 10 3 N
A ~ 75xl0“ 6 m 2
= 0.2 x 10 9 Pa = 200 MPa
A wire of circular cross-section, has
a tensile force of 60.0 N applied to it and this
force produces a stress of 3.06 MPa in the wire.
Determine the diameter of the wire.
Force F- 60.0 N and stress o - 3.06 MPa
= 3.06 x 10 6 Pa
0 . F F 60.ON
Since (J = ~j then area, A = — = -
A o 3.06 x 10 6 Pa
= 19.61 x 10 _6 m 2 = 19.61 mm 2
Figure 3.4
(a) For the crane, A, a supporting member, is in com¬
pression, B , a horizontal beam, is in shear, and
C, a rope, is in tension.
(b) For the transmission joint, parts D and F are in
tension, and E, the rivet or bolt, is in shear.
3.5 Stress
Forces acting on a material cause a change in dimensions
and the material is said to be in a state of stress. Stress
is the ratio of the applied force F to cross-sectional
area A of the material. The symbol used for tensile and
compressive stress is o (Greek letter sigma). The unit of
stress is the pascal*, Pa, where 1 Pa = 1 N/m 2 . Hence
F n
a- — Pa
A
where F is the force in newtons and A is the cross-sec¬
tional area in square metres. For tensile and compres¬
sive forces, the cross-sectional area is that which is at
right angles to the direction of the force. For a shear
F
force the shear stress is equal to —, where the cross-
sectional area A is that which is parallel to the direction
of the shear force. The symbol used for shear stress is
the Greek letter tau, r.
Cross-sectional area
hence
A =
19.61 =
nd 2
"T"
nd 2
"T"
* Blaise Pascal (19 June 1623-19 August 1662) was a French
polymath. A child prodigy, he wrote a significant treatise on
the subject of projective geometry at the age of sixteen, and
later corresponded with Pierre de Fermat on probability theory,
strongly influencing the development of modem economics
and social science.
To find out more go to www.routledge.com/cw/bird
Part Two
Part Two
50 Mechanical Engineering Principles
from which,
, 2= 4x1^1
K
and
i.e.
4 x 19.61
K
diameter of wire = 5.0 mm
Now try the following Practice Exercise
Practice Exercise 21 Further problems
on stress
1. A rectangular bar having a cross-sectional
area of 80 mm 2 has a tensile force of 20 kN
applied to it. Determine the stress in the bar.
[250 MPa]
2. A circular section cable has a tensile force of
1 kN applied to it and the force produces a
stress of 7.8 MPa in the cable. Calculate the
diameter of the cable. [12.78 mm]
3. A square-sectioned support of side 12 mm is
loaded with a compressive force of 10 kN.
Determine the compressive stress in the sup¬
port. [69.44 MPa]
4. A bolt having a diameter of 5 mm is load¬
ed so that the shear stress in it is 120 MPa.
Determine the value of the shear force on the
bolt. [2.356 kN]
5. A split pin requires a force of 400 N to shear
it. The maximum shear stress before shear
occurs is 120 MPa. Determine the minimum
diameter of the pin. [2.06 mm]
6 . A tube of outside diameter 60 mm and in¬
side diameter 40 mm is subjected to a tensile
load of 60 kN. Determine the stress in the
tube. [38.2 MPa]
3.6 Strain
The fractional change in a dimension of a material pro¬
duced by a force is called the strain. For a tensile or
compressive force, strain is the ratio of the change of
length to the original length. The symbol used for strain
is s (Greek epsilon). For a material of length L metres
which changes in length by an amount x metres when
subjected to stress,
x
8 = —
L
Strain is dimension-less and is often expressed as a
percentage, i.e.
x
percentage strain = — x 100
For a shear force, strain is denoted by the symbol
y (Greek letter gamma) and, with reference to Figure 3.5,
is given by:
y
X
L
Applied
shear force
Reaction
force
Figure 3.5
A bar 1.60 m long contracts axially
by 0.1 mm when a compressive load is applied to
it. Determine the strain and the percentage strain.
. contraction 0.1mm
Strain £ = -=-
original length l .60 x 10 3 mm
—= 0.0000625
1600
Percentage strain = 0.0000625 x 100 = 0.00625%
A wire of length 2.50 m has a per¬
centage strain of 0.012% when loaded with a
tensile force. Determine the extension of the wire.
Original length of wire = 2.50 m = 2500 mm
0.012
and strain = “[qq - -0.00012
Strain
extension
£ =
extension*
hence,
original length L
x = £L = (0.00012)(2500) = 0.30 mm
(a) A rectangular section metal bar
has a width of 10 mm and can support a maximum
compressive stress of 20 MPa; determine the mini¬
mum breadth of the bar when loaded with a force
of 3 kN. (b) If the bar in (a) is 2 m long and decreas¬
es in length by 0.25 mm when the force is applied,
determine the strain and the percentage strain.
The effects of forces on materials 51
^. force F
Since stress, o- -
area ^4
, , F 3000N
then, area, A- — =-
o 20 x 10 6 Pa
= 150 x 10 -6 m 2 = 150 mm 2
Cross-sectional area = width x breadth, hence
area 150
breadth = ——r = ttt = 15 mm
width 10
. contraction 0.25
(b) Strain, £ = ————-- = —— = 0.000125
original length 2000
Percentage strain = 0.000125 x 100 = 0.0125%
A pipe has an outside diameter of
25 mm, an inside diameter of 15 mm and length
0.40 m and it supports a compressive load of
40 kN. The pipe shortens by 0.5 mm when the load
is applied. Determine (a) the compressive stress
(b) the compressive strain in the pipe when
supporting this load.
Compressive force F = 40 kN = 40000 N,
and cross-sectional area^4 = — |.D 2 ~ d 2
where D = outside diameter = 25 mm
and d = inside diameter =15 mm.
Hence, A = —(25 2 - 15 2 )mm 2
4
= —(25 2 - 15 2 ) x 10 _ 6 m 2
4
= 3.142 xl0" 4 m 2
F 40000N
(a) Compressive stress, o - — — - - —-
A 3.142 x 10 -4 m 2
= 12.73 x 10 7 Pa
= 127.3 MPa
(b) Contraction of pipe when loaded,
x = 0.5 mm = 0.0005 m, and original length
L - 0.40 m. Hence, compressive strain,
8 -
X
T
0.0005
0.4
= 0.00125 (or 0.125%)
A circular hole of diameter 50 mm is
to be punched out of a 2 mm thick metal plate. The
shear stress needed to cause fracture is 500 MPa.
Determine (a) the minimum force to be applied
to the punch, and (b) the compressive stress in the
punch at this value.
(a) The area of metal to be sheared, A = perimeter of
hole x thickness of plate.
Perimeter of hole = nd = 7r(50 x 10 3 ) = 0.1571 m.
Hence, shear area, ^4 = 0.1571 x2x 10 ~ 3
= 3.142 x 10 -4 m 2
Since shear stress =
force
area
shear force = shear stress x area
= (500 x 10 6 x 3.142 x 10~ 4 )N
= 157.1 kN,
which is the minimum force to be applied to the punch,
(b) Are, of punch = ^
= 0.001963 m 2
Compressive stress
_ force _ 157.1 x 10 3 N
area 0.001963m 2
= 8.003 x 10 7 Pa = 80.03 MPa,
which is the compressive stress in the punch.
A rectangular block of plastic
material 500 mm long by 20 mm wide by 300 mm
high has its lower face glued to a bench and a force
of 200 N is applied to the upper face and in line
with it. The upper face moves 15 mm relative to
the lower face. Determine (a) the shear stress, and
(b) the shear strain in the upper face, assuming the
deformation is uniform.
force
(a) Shear stress, r =-——:— -
area parallel to the torce
Area of any face parallel to the force
= 500 mm x 20 mm
= (0.5 x 0.02) m 2 = 0.01 m 2
200N
Hence, shear stress, r =- -
0.01m 2
= 20000 Pa or 20 kPa
x
(b) Shear strain, y - — (see side view in Figure 3.6)
15
300
= 0.05 (or 5%)
Part Two
Part Two
52 Mechanical Engineering Principles
Applied
shear force
L = 300 mm
—
\
I
l
i
i
yi
1
1
m
< i
i
i
i
i
i
i
1
1
1
1
1
1
1
1
1
i
i
1
1
Reaction
force
Figure 3.6
and in line with it. The upper face moves
12 mm relative to the lower face. Determine
(a) the shear stress, and (b) the shear strain in
the upper face, assuming the deformation is
uniform. [(a) 25 kPa (b) 0.04 or 4%]
3.7 Elasticity, limit of proportionality
and elastic limit
Now try the following Practice Exercise
Practice Exercise 22 Further problems
on strain
1. A wire of length 4.5 m has a percentage
strain of 0.050% when loaded with a tensile
force. Determine the extension in the wire.
[2.25 mm]
2. A metal bar 2.5 m long extends by 0.05 mm
when a tensile load is applied to it. Deter¬
mine (a) the strain, (b) the percentage strain.
[(a) 0.00002 (b) 0.002%]
3. An 80 cm long bar contracts axially by
0.2 mm when a compressive load is applied
to it. Determine the strain and the percentage
strain. [0.00025,0.025%]
4. A pipe has an outside diameter of 20 mm,
an inside diameter of 10 mm and length
0.30 m and it supports a compressive load of
50 kN. The pipe shortens by 0.6 mm when
the load is applied. Determine (a) the com¬
pressive stress, (b) the compressive strain in
the pipe when supporting this load.
[(a) 212.2 MPa (b) 0.002 or 0.20%]
Elasticity is the ability of a material to return to its orig¬
inal shape and size on the removal of external forces.
Plasticity is the property of a material of being perma¬
nently deformed by a force without breaking. Thus if a
material does not return to the original shape, it is said
to be plastic.
Within certain load limits, mild steel, copper, poly¬
thene and rubber are examples of elastic materials; lead
and plasticine are examples of plastic materials.
If a tensile force applied to a uniform bar of mild steel
is gradually increased and the corresponding extension of
the bar is measured, then provided the applied force is not
too large, a graph depicting these results is likely to be as
shown in Figure 3.7. Since the graph is a straight line,
extension is directly proportional to the applied force.
Figure 3.7
5. When a circular hole of diameter 40 mm is
punched out of a 1.5 mm thick metal plate,
the shear stress needed to cause fracture is
100 MPa. Determine (a) the minimum force
to be applied to the punch, and (b) the com¬
pressive stress in the punch at this value.
[(a) 18.85 kN (b) 15.0 MPa]
6 . A rectangular block of plastic material
400 mm long by 15 mm wide by 300 mm
high has its lower face fixed to a bench and
a force of 150 N is applied to the upper face
The point on the graph where extension is no longer
proportional to the applied force is known as the limit
of proportionality. Just beyond this point the mate¬
rial can behave in a non-linear elastic manner, until the
elastic limit is reached. If the applied force is large, it
is found that the material becomes plastic and no longer
returns to its original length when the force is removed.
The material is then said to have passed its elastic limit
and the resulting graph of force/extension is no longer
F
a straight line. Stress, o -—, from Section 3.5, and
A
since, for a particular bar, area A can be considered as
a constant, then F a o.
The effects offerees on materials 53
X
Strain £ = —, from Section 3.6, and since for a particu¬
lar bar L is constant, then x as. Hence for stress applied
to a material below the limit of proportionality a graph
of stress/strain will be as shown in Figure 3.8, and is a
similar shape to the force/extension graph of Figure 3.7.
Figure 3.8
3.8 Hooke's law
Hooke’s law states:
Within the limit of proportionality , the extension of a
material is proportional to the applied force
It follows, from Section 3.7, that:
Within the limit of proportionality of a material, the
strain produced is directly proportional to the stress
producing it
Young’s modulus of elasticity
Within the limit of proportionality, stress a strain, hence
stress = (a constant) x strain
This constant of proportionality is called Young’s
modulus of elasticity* and is given the symbol E.
The value of E may be determined from the gradient
of the straight line portion of the stress/strain graph.
The dimensions of E are pascals (the same as for stress,
since strain is dimension-less).
a
E= — Pa
£
Some typical values for Young’s modulus of elasticity,
E , include:
Aluminium alloy 70 GPa (i.e. 70 x 10 9 Pa), brass
90 GPa, copper 96 GPa, titanium alloy 110 GPa,
diamond 1200 GPa, mild steel 210 GPa, lead 18 GPa,
tungsten 410 GPa, cast iron 110 GPa, zinc 85 GPa, glass
fibre 72 GPa, carbon fibre 300 GPa.
Stiffness
A material having a large value of Young’s modulus is
said to have a high value of material stiffness, where
stiffness is defined as:
Stiffness =
force F
extension x
For example, mild steel is a much stiffer material than
x
L
lead.
(7
F
Since
E =
, ( 7 = —
£
A
F
then
E =
J_
FL
X
Ax
/
F\L
x 1 A
L
i.e. E - (stiffness) x
/ \
F
v
/ \
L
~A
v
\
7
V X 7
is also the gradient of the force/exten-
Stiffness
sion graph, hence
E = (gradient of force/extension graph)
V
* homas Young (13 June 1773 - 10 May 1829) was an English
polymath. He is famous for having partly deciphered Egyptian
hieroglyphics (specifically the Rosetta Stone). Young made
notable scientific contributions to the fields of vision, light, solid
mechanics, energy, physiology, language, musical harmony and
Egyptology. To find out more go to www.routledge.com/cw/bird
Part Two
Part Two
54 Mechanical Engineering Principles
Since L and A for a particular specimen are constant,
the greater Young’s modulus the greater the material
stiffness.
A wire is stretched 2 mm by a force
of 250 N. Determine the force that would stretch
the wire 5 mm, assuming that the limit of propor¬
tionality is not exceeded.
Hooke’s law states that extension x is proportional to
force F, provided that the limit of proportionality is not
exceeded, i.e. x a F or x = kF where A: is a constant.
When x — 2 mm, F = 250 N,
thus 2 = £(250), from which,
constant k =-=-
250 125
When x = 5 mm, then 5 = kF
/ \
i.e.
5 =
1
v !25 y
F
from which, force F= 5(125) = 625 N
Thus to stretch the wire 5 mm, a force of 625 N is
required.
A tensile force of 10 kN applied to a
component produces an extension of 0.1 mm.
Determine (a) the force needed to produce an ex¬
tension of 0.12 mm, and (b) the extension when
the applied force is 6 kN, assuming in each case
that the limit of proportionality is not exceeded.
From Hooke’s law, extension x is proportional to
force F within the limit of proportionality, i.e. x a F or
x = kF , where k is a constant. If a force of 10 kN
produces an extension of 0.1 mm, then 0.1 = £( 10 ),
from which, constant k- — = 0.01
10
(a) When an extension x = 0.12 mm, then
0.12 = £(F), i.e. 0.12 = 0.01 F, from which,
0 12
force F = = 12 kN
0.01
(b) When force F - 6 kN, then
extension* = k{6) = (0.01)(6) = 0.06 mm
A copper rod of diameter 20 mm
and length 2.0 m has a tensile force of 5 kN
applied to it. Determine (a) the stress in the rod
(b) by how much the rod extends when the load is
applied. Take the modulus of elasticity for copper
as 96 GPa.
(a) Force F= 5 kN = 5000 N and cross-sectional area
A= 7r ^L= 7r ( Q - Q2Q ) = 0.000314 m 2
4 4
^ F 5000N
Stress, <r = — =-
A 0.000314m 2
= 15.92 x 10 6 Pa = 15.92 MPa
,,, c . r o . ■ o 15.92 xlO 6 Pa
(b) Since E - — then strain s - — =-
£ E 96 x 10 9 Pa
= 0.000166
Strain s = — , hence extension,
L
x = sL = (0.000166)(2.0) = 0.000332 m
i.e. extension of rod is 0.332 mm
A bar of thickness 15 mm and hav¬
ing a rectangular cross-section carries a load of
120 kN. Determine the minimum width of the bar
to limit the maximum stress to 200 MPa. The bar,
which is 1.0 m long, extends by 2.5 mm when
carrying a load of 120 kN. Determine the modulus
of elasticity of the material of the bar.
Force, F - 120 kN = 120000 N and cross-sectional area
A = (15x)10 -6 m 2 , where x is the width of the rectangu¬
lar bar in millimetres.
Stress
o- —, from which,
A
F 120000 N . 1A _ 4
o 200 x 10 6 Pa
= 6x 10 -4 x 10 6 mm 2
= 6 x 10 2 mm 2 = 600 mm 2
Hence,
600 = 15x, from which,
width of bar, * =
600
~TT
= 40 mm
Extension of bar = 2.5 mm = 0.0025 m
0 . x 0.0025
Strain s = — = -
L 1.0
= 0.0025
Modulus of elasticity, E =
stress _ 200 xlO 6
strain 0.0025
80 x 10 9 = 80 GPa
An aluminium alloy rod has a length
of 200 mm and a diameter of 10 mm. When sub¬
jected to a compressive force the length of the rod is
The effects offerees on materials 55
199.6 mm. Determine (a) the stress in the rod when
loaded, and (b) the magnitude of the force. Take the
modulus of elasticity for aluminium alloy as 70 GPa.
(a) Original length of rod, L = 200 mm, final length of
rod = 199.6 mm, hence contraction, x = 0.4 mm.
Thus, strain, s = — = = 0.002
L 200
Modulus of elasticity, E - St - esS - (J -
strain £
hence stress, o = Es = 70 x 10 9 x 0.002
= 140 x 10 6 Pa = 140 MPa
0 . force F
Since stress o- -
area A
then force, F - oA
Cross-sectional area, A - 71
4 4
= 7.854 xlO" 5 m 2
Hence, compressive force,
F -oA- 140 x 10 6 x 7.854 x 10~ 5 = 11.0 kN
A brass tube has an internal diameter
of 120 mm and an outside diameter of 150 mm and
is used to support a load of 5 kN. The tube is 500 mm
long before the load is applied. Determine by how
much the tube contracts when loaded, taking the
modulus of elasticity for brass as 90 GPa.
Force in tube, F - 5 kN = 5000 N, and
cross-sectional area of tube,
A = ^[D 2 - J 2 ) = ^(0.150 2 - 0.120 2 )
= 0.006362m 2
Stress in tube, o
F _ 5000N
A 0.006362m 2
= 0.7859 x 10
Since the modulus of elasticity, E =
6 Pa
stress <7
strain £
then strain, s = —
E
0.7859 x 10 6 Pa
90 x 10 9 Pa
= 8.732 x 1(T 6
Thus, when loaded, the tube contracts by 4.37 pm
In an experiment to determine the
modulus of elasticity of a sample of mild steel,
a wire is loaded and the corresponding extension
noted. The results of the experiment are as shown.
Load (N) 0 40 110 160 200 250 290 340
Extension 0 1.2 3.3 4.8 6.0 7.5 10.0 16.2
(mm)
Draw the load/extension graph.
The mean diameter of the wire is 1.3 mm and
its length is 8.0 m. Determine the modulus of
elasticity E of the sample, and the stress at the limit
of proportionality.
A graph of load/extension is shown in Figure 3.9
360
320
280
240
z
1 200
o
" J 160
120
80
40
4
0
Figure 3.9
l']L)
{ x X A J
— is the gradient of the straight line part of the load/
x
extension graph.
<7
E= —
£
F
A
X
L
D
B
C
6 8 10 12 14 16 18
Extension/mm
Strain, s =
contraction, x =
contraction x
thus,
original length L
sL — 8.732 x 10 —6 x 0.500
= 4.37 x 10 -6 m.
Gradient,
F
x
BC
AC
200N
-— = 33.33
6x10 3 m
Modulus of elasticity = (gradient of graph)
x 10 3 N/m
r L
A
v /
Part Two
Part Two
56 Mechanical Engineering Principles
Length of specimen, L - 8.0 m and cross-sectional area
{ _Kd * 1 2 _ tt(0.0013) 2 * *
4 4
= 1.327 x 10" 6 m 2
Hence modulus of elasticity,
/
E = (33.33 x 10 3 )
8.0
\
v
1.327 xlO" 6
= 201 GPa
7
The limit of proportionality is at point D in Fig¬
ure 3.9 where the graph no longer follows a straight
line. This point corresponds to a load of 250 N as
shown.
4. A test to determine the load/extension
graph for a specimen of copper gave the
following results:
Load (kN)
8.5
15.0
23.5
30.0
Extension (mm)
0.04
0.07
0.11
0.14
Plot the load/extension graph, and from
the graph determine (a) the load at an ex¬
tension of 0.09 mm, and (b) the extension
corresponding to a load of 12.0 kN.
[(a) 19 kN (b) 0.057 mm]
Stress at the limit of proportionality
_ force _ 250
area 1.327 x 10” 6
= 188.4 x 10 6 Pa = 188.4 MPa
Note that for structural materials the stress at the elastic
limit is only fractionally larger than the stress at the
limit of proportionality, thus it is reasonable to assume
that the stress at the elastic limit is the same as the
stress at the limit of proportionality; this assumption
is made in the remaining exercises. In Figure 3.9, the
elastic limit is shown as point F.
Now try the following Practice Exercise
Practice Exercise 23 Further problems on
Hooke's law
1. A wire is stretched 1.5 mm by a force of
300 N. Determine the force that would
stretch the wire 4 mm, assuming the elastic
limit of the wire is not exceeded.
[800 N]
2. A rubber band extends 50 mm when a force
of 300 N is applied to it. Assuming the
band is within the elastic limit, determine
the extension produced by a force of 60 N.
[10 mm]
3. A force of 25 kN applied to a piece of
steel produces an extension of 2 mm.
Assuming the elastic limit is not exceeded,
determine (a) the force required to produce
an extension of 3.5 mm, (b) the extension
when the applied force is 15 kN.
[(a) 43.75 kN (b) 1.2 mm]
5. A circular section bar is 2.5 m long and has
a diameter of 60 mm. When subjected to a
compressive load of 30 kN it shortens by
0.20 mm. Determine Young’s modulus of
elasticity for the material of the bar.
[132.6 GPa]
6 . A bar of thickness 20 mm and having a
rectangular cross-section carries a load of
82.5 kN. Determine (a) the minimum width
of the bar to limit the maximum stress to
150 MPa, (b) the modulus of elasticity of
the material of the bar if the 150 mm long
bar extends by 0.8 mm when carrying a
load of 200 kN.
[(a) 27.5 mm (b) 68.2 GPa]
7. A metal rod of cross-sectional area
100 mm 2 carries a maximum tensile load
of 20 kN. The modulus of elasticity for
the material of the rod is 200 GPa. Deter¬
mine the percentage strain when the rod is
carrying its maximum load. [0.10%]
8 . A metal tube 1.75 m long carries a tensile
load and the maximum stress in the tube
must not exceed 50 MPa. Determine the
extension of the tube when loaded if the
modulus of elasticity for the material is
70 GPa. [1.25 mm]
9. A piece of aluminium wire is 5 m long and
has a cross-sectional area of 100 mm 2 . It is
subjected to increasing loads, the extension
being recorded for each load applied. The
results are:
Load (kN)
0
1.12
2.94
4.76
7.00
9.10
Extension
(mm)
0
0.8
2.1
3.4
5.0
6.5
The effects offerees on materials 57
Draw the load/extension graph and hence
determine the modulus of elasticity for the
material of the wire. [70 GPa]
10. In an experiment to determine the modulus
of elasticity of a sample of copper, a wire
is loaded and the corresponding extension
noted. The results are:
Load (N)
0
20
34
72
94
120
Extension
(mm)
0
0.7
1.2
2.5
3.3
4.2
Draw the load/extension graph and de¬
termine the modulus of elasticity of the
sample if the mean diameter of the wire is
1.23 mm and its length is 4.0 m.
[96 GPa]
3.9 Ductility, brittleness and
malleability
Ductility is the ability of a material to be plastically
deformed by elongation, without fracture. This is a
property that enables a material to be drawn out into
wires. For ductile materials such as mild steel, copper
and gold, large extensions can result before fracture
occurs with increasing tensile force. Ductile materials
usually have a percentage elongation value of about
15% or more.
is the property of a material manifested
by fracture without appreciable prior plastic defor¬
mation. Brittleness is a lack of ductility, and brittle
materials such as cast iron, glass, concrete, brick and
ceramics, have virtually no plastic stage, the elastic
stage being followed by immediate fracture. Little or
no ‘waist’ occurs before fracture in a brittle material
undergoing a tensile test.
Malleability is the property of a material whereby
it can be shaped when cold by hammering or rolling.
A malleable material is capable of undergoing plastic
deformation without fracture. Examples of malleable
materials include lead, gold, putty and mild steel.
Sketch typical load/extension
curves for (a) an elastic non-metallic material,
(b) a brittle material and (c) a ductile material.
Give a typical example of each type of material.
(a) A typical load/extension curve for an elastic non-
metallic material is shown in Figure 3.10(a), and
an example of such a material is polythene.
Figure 3.10
(b) A typical load/extension curve for a brittle ma¬
terial is shown in Figure 3.10(b), and an example
of such a material is cast iron.
(c) A typical load/extension curve for a ductile ma¬
terial is shown in Figure 3.10(c), and an example
of such a material is mild steel.
3.10 Modulus of rigidity
Experiments have shown that under pure torsion (see
Chapter 12), up to the limit of proportionality, we have
Hooke’s law in shear, where
shear stress
shear strain
= rigidity (or shear) modulus
or -=G (3.1)
7
where r = shear stress, y = shear strain (see Figures 3.5
and 3.6) and G - rigidity (or shear) modulus.
3.11 Thermal strain
If a bar of length L and coefficient of linear expansion a
were subjected to a temperature rise of J, its length will
increase by a distance aLT\ as described in Chapter 22.
Part Two
Part Two
58 Mechanical Engineering Principles
Thus the new length of the bar will be:
L + clLT = L( 1 + aT)
Now, as the original length of the bar was T, then the
thermal strain due to a temperature rise will be:
extension aLT
s - ——-=-
original length L
i.e. thermal strain, s = aT
However, if the bar were not constrained, so that it can
expand freely, there will be no thermal stress.
If, however, the bar were prevented from expanding
then there would be a compressive stress in the bar.
Now e _ original length - new length
original length
_ L - L( 1 + aT)
~ Z
_ L - L - LaT
Z
i.e. strain, s = -aT
and, since stress = strain x E,
then a — — aTE
= 2 x 10 11 N/m 2 x (- 210 x 10“ 6 )
i.e. cr r =-42MPa
Hence, the stress at 35°C = initial stress + a T
= (-30 - 42) MPa
i.e. <7 = -72 MPa
(b) For the prop to be ineffective, it will be neces¬
sary for the temperature to fall so that there is no
stress in the prop, that is, from 20°C the tempera¬
ture must fall so that the initial stress of 30 MPa
is nullified. Hence, drop in stress = -30 MPa
Therefore, drop in thermal strain
= — = - 3Qxl ° 6pa =_ 1.5 x 10~ 4
E 2 x 10 11 Pa
Thermal strain, s = aT
from which, temperature
T _ e _ -1.5x10 4 __ jo 7 °c
a 14 x 10 -6
Hence, the drop in temperature T from 20°C is
-t m r\ s—i
Therefore, the temperature for the prop to be
ineffective = 20°C - 10.7°C = 9.3°C
A steel prop is used to stabilise a
building, as shown in Figure 3.11. (a) If the
compressive stress in the bar at 20°C is 30 MPa,
what will be the stress in the prop if the
temperature is raised to 35°C? (b) At what tem¬
perature will the prop cease to be effective?
Take E = 2 x 10 n N/m 2 and a = 14 x 10“ 6 /°C.
(a) Additional thermal strain, s T = -aT
= -(14 x 10- 6 /°C) x (35 - 20)°C
i.e. £j, — —14 x 10 _6 x 15 = —210 x 10 —6
Additional thermal stress, a T - Es r
Now try the following Practice Exercise
Practice Exercise 24 Further problem on
thermal strain
1. A steel rail may be assumed to be stress free
at 5°C. If the stress required to cause buck¬
ling of the rail is -50 MPa, at what tempera¬
ture will the rail buckle? It may be assumed
that the rail is rigidly fixed at its ‘ends’. Take
E = 2x 10 n N/m 2 and a = 14 x 10- 6 / o C.
[22.86°C]
3.12 Compound bars
Compound bars are of much importance in a number of
different branches of engineering, including reinforced
concrete pillars, composites, bimetallic bars, and so on.
In this section, solution of such problems usually in¬
volves two important considerations, namely
(a) compatibility (or considerations of displace¬
ments)
(b) equilibrium
The effects of forces on materials 59
N.B. It is necessary to introduce compatibility in
this section as compound bars are, in general, statically
indeterminate (see Chapter 7). The following worked
problems demonstrate the method of solution.
A solid bar of cross-sectional area
A j, Young’s modulus E { and coefficient of linear
expansion cq is surrounded co-axially by a hollow
tube of cross-sectional area A 2 , Young’s modulus
E 2 and coefficient of linear expansion a 2 , as shown
in Figure 3.12. If the two bars are secured firmly
to each other, so that no slipping takes place with
temperature change, determine the thermal stresses
due to a temperature rise T. Both bars have an
initial length L and cq > a 2
Figure 3.12 Compound bar
There are two unknown forces in these bars, namely F {
and F 2 ; therefore, two simultaneous equations will be
required to determine these unknown forces. The first
equation can be obtained by considering the compat¬
ibility (i.e. ‘deflections’) of the bars, with the aid of
Figure 3.13.
A
A
Figure 3.13 ‘Deflections’ of compound bar
Free expansion of bar (1) = cq LT
Free expansion of bar (2) = a 2 LT
In practice, however, the final resting position of the
compound bar will be somewhere in between these two
positions (i.e. at the position A-A in Figure 3.13). To
achieve this, it will be necessary for bar ( 2 ) to be pulled
out by a distance e 2 L and for bar (2) to be pushed in by
a distance s { L , where
£j = compressive strain in ( 1 )
and s 2 = tensile strain in ( 2 )
From considerations of compatibility (‘deflection’) in
Figure 3.13,
a x LT — s x L — a 2 LT + s 2 L
Dividing throughout by L gives:
a { T-s 1 = a 2 T + s 2
and s 1 = ((Xy —a 2 )T—s 2
Now, (T [ =E l £ {
and o 2 - E 2 s 2 (or e 2 =
- °2 )
E
2
Hence, cq - (cq - a 2 ) E { T-E x s 2
I?
i.e. cq = (cq - a ? ) E { T-o 2 —1
E
(3.2)
To obtain the second simultaneous equation, it will be
necessary to consider equilibrium of the compound bar.
Let F { = unknown compressive force in bar (1)
and F 2 - unknown tensile force in bar (2)
Now, from equilibrium considerations,
f { = f 2
F \ F
but a, = — and =
l
A
l
Ar
Therefore,
Ml ~ ^2^2
or
°i =
G 2 A2
(3.3)
Equating equations (3.2) and (3.3) gives
(7 2 A 2 E }
=(.<*!-a 2 )EJ-CT 2 —
E,
, o —I , °2 ^2
i.e. ^ +
2E
2
(
A
l
and & 2
F i + ^2
v 2
A
i
from which,
= (a { - a^E^T
= (oq — a 2 )E { T
(a { - ot 2 )E x T
<t 2 =
F\ + ^2
V ^2
A,
1 2
(a x -a 2 )EJ
\e 1+ a 2 eT
\
e 2 a x
7
Part Two
Part Two
60 Mechanical Engineering Principles
i.e. d =
and a, =
(ot x oc 2 )E 1 E 2 A x T
(A\ E j + A 2 E 2 )
(tensile)
(3.4)
G 2 A-2
1
A
l
i.e.
<7i =
_ ( cc j a 2 )E x E 2 A 2 T
(A x E x + A 2 E 2 )
(compressive)
(3.5)
If the solid bar of Problem 19 did
not suffer temperature change, but instead was
subjected to a tensile axial force P, as shown in
Figure 3.14, determine g { and g 2
< - L ->
Figure 3.14 Compound bar under axial tension
There are two unknown forces in this bar, namely, F {
and F 2 ; therefore, two simultaneous equations will be
required.
The first of these simultaneous equations can be ob¬
tained by considering compatibility, i.e.
or
But
Therefore,
or
Now,
Hence,
or
deflection of bar ( 1 ) = deflection of bar ( 2 )
d l= d 2
d x = e x L and d 2 = s 2 L
t, X L — l 2 L
£ \ = S 2
G
a
l
d
E x and e 2 = t
do
E
l
Er
G 2 E\
a \ = 21
E.
(3.6)
The second simultaneous equation can be obtained by
considering the equilibrium of the compound bar.
Let F { = tensile force in bar (1)
and F ? = tensile force in bar ( 2 )
Now, from equilibrium conditions
p=f 1 + f 2
i.e. P = g ] A x +g 2 A 2 (3.7)
Substituting equation (3.6) into equation (3.7) gives:
g 2 E\
—j? A 1 +
(EiA {
/
\
+ a 2
E x + A 2 E 2
\
/
PE 2
Rearranging gives: * 2 = ^ ^ + ^ ^ (3-8)
Since
then
°2^\
E 2
PE,
i (A, E x + A 2 E 2 )
(3-9)
N.B. If P is a compressive force, then both g { and g 2
will be compressive stresses (i.e. negative), and vice-
versa if P were tensile.
A concrete pillar, which is rein¬
forced with steel rods, supports a compressive
axial load of 2 MN.
(a) Determine stresses g { and g 2 given the fol¬
lowing:
For the steel, A l = 4 x 10 -3 m 2 and
E x =2x 10 11 N/m 2
For the concrete, 4 2 = 0.2m 2 and
E 2 = 2x 10 10 N/m 2
(b) What percentage of the total load does the
steel reinforcement take?
(a) From equation (3.9),
a PE '
1 (A, E 1 + A 2 E 2 )
2 x 10 6 x 2 x 10 11
“ ~ (4 x 10“ 3 x 2 x 10 11 + 0.2 x 2 x 10 10 )
4x 10 17 _ 4x 10 17 _ 10 9
(8 x 10 8 + 40 x 10 8 ) 48 x 10 8 12
= -83.3 x 10 6
i.e. the stress in the steel,
oq = - 83.3 MPa (3.10)
From equation (3.8),
PE 2
CT2 ~~(A l E l + A 2 E 2 )
2 x 10 6 x 2 x 10 10
(4 x KT 3 x 2 x 10 11 + 0.2 x 2 x 10 10 )
The effects offerees on materials 61
4 x to 16 _ 4 x 10 16 _ 10 8
(8 x 10 8 + 40 x 10 8 ) 48 x 10 8 12
= -8.3 x 10 6
i.e. the stress in the concrete,
a 2 = -8.3 MPa (3.11)
Force in the steel,
F \=°\ A \
= - 83.3 x 10 6 x 4 x 10- 3 = 3.33 X 10 5 N
Therefore, the percentage total load taken by
the steel reinforcement
=--xl00%
total axial load
3.33 x 10 5
2 x 10 6
x 100% = 16.65%
If the pillar of problem 21 were
subjected to a temperature rise of 25°C, what
would be the values of stresses a { and <r 2 ?
Assume the coefficients of linear expansion are,
for steel, a l = 14 x 10 _6 /°C, and for concrete,
a 2 = 12 x 10~ 6 /°C.
As is larger than a 2 , the effect of a temperature rise
will cause the Thermal stresses’ in the steel to be com¬
pressive and those in the concrete to be tensile.
From equation (3.5), the thermal stress in the steel,
(a { a 2 )E l E 2 A 2 T
(^i + A 2 E 2 )
(l4xl0- 6 -12xl0“ 6 )
x 2 x 10 11 x 2 x 10 10 x 0.2 x 25
48 x 10 8
40 x 10 15
48 x 10 8
= -0.833 x 10 7 = -8.33 x 10 6
= -8.33 MPa (3.12)
From equation (3.3), the thermal stress in the concrete,
< 7 l A l (-8.33 x 10 6 ) x 4 x 10 -3
2 A 2 0.2
= 0.167 MPa (3.13)
From equations (3.10) to (3.13):
<7 X = -83.3 - 8.33 = - 91.63 MPa
and <t 2 = -8.3 + 0.167 = -8.13 MPa
Now try the following Practice Exercises
Practice Exercise 25 Further problems on
compound bars
1. Two layers of carbon fibre are stuck to each
other, so that their fibres he at 90° to each
other, as shown in Figure 3.15. If a tensile
force of 1 kN were applied to this two-layer
compound bar, determine the stresses in each.
For layer 1, E x = 300 GPa and^j = 10 mm 2
For layer 2, E 2 = 50 GPa and A ? = A l =
10 mm 2
Layer 1
Layer 2
Figure 3.15 Carbon fibre layers
[cr 1 = 85.71 MPa, a 2 = 14.29 MPa]
2. If the compound bar of Problem 1 were sub¬
jected to a temperature rise of 25°C, what
would the resulting stresses be? Assume
the coefficients of linear expansion are, for
layer 1, a l = 5 x 10 -6 /°C, and for layer 2,
a 2 = 0.5 x 10- 6 /°C.
[cj { = 80.89 MPa, o 2 - 19.11 MPa]
Practice Exercise 26 Short-answer
questions on the
effects of forces on
materials
1. Name three types of mechanical force that
can act on a body.
2. What is a tensile force? Name two practical
examples of such a force.
3. What is a compressive force? Name two
practical examples of such a force.
4. Define a shear force and name two practi¬
cal examples of such a force.
5. Define elasticity and state two examples of
elastic materials.
6 . Define plasticity and state two examples of
plastic materials.
Part Two
Part Two
62 Mechanical Engineering Principles
7. Define the limit of proportionality.
8 . State Hooke’s law.
9. What is the difference between a ductile
and a brittle material?
10. Define stress. What is the symbol used for
(a) a tensile stress (b) a shear stress?
11. Strain is the ratio — •
12. The ratio -— is called.
strain
13. State the units of (a) stress (b) strain
(c) Young’s modulus of elasticity.
14. Stiffness is the ratio —.
15. Sketch on the same axes a typical load/
extension graph for a ductile and a brittle
material.
16. Define (a) ductility (b) brittleness (c) mal¬
leability.
17. Define rigidity modulus.
18. The new length L 2 of a bar of length
of coefficient of linear expansion a,
when subjected to a temperature rise T is:
19. The thermal strain s due to a temperature
rise T in material of coefficient of linear ex¬
pansion a is given by: s = .
Practice Exercise 27 Multiple-choice
questions on the
effects of forces on
materials
(Answers on page 335)
1. The unit of strain is:
(a) pascals (b) metres
(c) dimension-less (d) newtons
2. The unit of stiffness is:
(a) newtons (b) pascals
(c) newtons per metre (d) dimension-less
3. The unit of Young’s modulus of elasticity is:
(a) pascals (b) metres
(c) dimension-less (d) newtons
4. A wire is stretched 3 mm by a force of
150 N. Assuming the elastic limit is not
exceeded, the force that will stretch the
wire 5 mm is:
(a) 150 N (b) 250 N
(c) 90 N (d) 450 N
5. For the wire in question 4, the extension
when the applied force is 450 N is:
(a) 1 mm (b) 3 mm
(c) 9 mm (d) 12 mm
6 . Due to the forces acting, a horizontal beam
is in:
(a) tension (b) compression
(c) shear
7. Due to forces acting, a pillar supporting a
bridge is in:
(a) tension (b) compression
(c) shear
8 . Which of the following statements is false?
(a) Elasticity is the ability of a material to
return to its original dimensions after
deformation by a load.
(b) Plasticity is the ability of a material to
retain any deformation produced in it
by a load.
(c) Ductility is the ability to be permanent¬
ly stretched without fracturing.
(d) Brittleness is the lack of ductility and a
brittle material has a long plastic stage.
9. A circular rod of cross-sectional area
100 mm 2 has a tensile force of 100 kN
applied to it. The stress in the rod is:
(a) 1 MPa (b) 1 GPa
(c) 1 kPa (d) 100 MPa
10. A metal bar 5.0 m long extends by 0.05 mm
when a tensile load is applied to it. The per¬
centage strain is:
(a) 0.1 (b) 0.01
(c) 0.001 (d) 0.0001
The effects offerees on materials 63
An aluminium rod of length 1.0 m and cross-sec¬
tional area 500 mm 2 is used to support a load of
5 kN which causes the rod to contract by 100 pm.
12. The strain in the rod.
13. Young’s modulus of elasticity.
For questions 11 to 13, select the correct answer
from the following list:
(a) 100 MPa
(b) 0.001
(c) lOkPa
(d) 100 GPa
(e) 0.01
(f) 10 MPa
(g) lOGPa
(h) 0.0001
(i) 10 Pa
11. The stress in the rod.
14. A compound bar of length L is subj ected to
a temperature rise of T. If > a 2 , the strain
in bar 1 will be:
(a) tensile (b) compressive
(c) zero (d) aT
15. For Problem 14, the stress in bar 2 will be:
(a) tensile (b) compressive
(c) zero (d) aT
For fully worked solutions to each of the problems in Practice Exercises 21 to 27 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Two
Chapter 4
Why it is important to understand: Tensile testing
In this chapter, additional material constants are introduced to help define the properties of materials.
In addition to this, a description is given of the standard tensile test used to obtain the strength of
various materials, especially the strength and material stiffness of metals. This is aided by a standard
tensile test, where the relationship of axial load on a specimen and its axial deflection are described. The
engineer can tell a lot from the results of this experiment. For example, the relationship between load
and deflection is very different for steel and aluminium alloy and most other materials, including copper,
zinc, brass, titanium, and so on. By carrying out this test, the engineer can determine the required
properties of different materials, and this method, together with its interpretation, is discussed in this
chapter. The importance in tensile testing is to know the failure criteria of various materials, because
different materials exhibit different properties in their strength, elasticity, etc.
At the end of this chapter you should be able to:
• describe a tensile test
• recognise from a tensile test the limit of proportionality, the elastic limit and the yield point
• plot a load/extension graph from given data
• calculate from a load/extension graph the modulus of elasticity, the yield stress, the ultimate tensile strength,
percentage elongation and the percentage reduction in area
• appreciate the meaning of proof stress
4.1 The tensile test
A tensile test is one in which a force is applied to a
specimen of a material in increments and the corre¬
sponding extension of the specimen noted. The pro¬
cess may be continued until the specimen breaks into
two parts and this is called testing to destruction. The
testing is usually carried out using a universal testing
machine that can apply either tensile or compressive
forces to a specimen in small, accurately measured
steps. British Standard 18 gives the standard proce¬
dure for such a test. Test specimens of a material are
made to standard shapes and sizes and two typical test
pieces are shown in Figure 4.1. The results of a tensile
test may be plotted on a load/extension graph and a
typical graph for a mild steel specimen is shown in
Figure 4.2.
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
Tensile testing 65
Diameter
i —
i
i
f-
Thickness
(b)
Figure 4.1
(v) For mild steel, the extension up to the point J is
some 40 times larger than the extension up to
the point B.
(vi) Shortly after point J, the material strain
hardens, where the slope of the load-extension
curve is about l/50th the slope of the
curve from A to B , for materials such as mild
steel.
(vii) Between points D and E extension takes
place over the whole gauge length of the
specimen.
(viii) Point E gives the maximum load which can
be applied to the specimen and is used to
determine the ultimate tensile strength (UTS)
of the specimen (often just called the tensile
strength)
maximum load
LJ lu — -
original cross-sectional area
(ix) Between points E and F the cross-sectional
area of the specimen decreases, usually about
half way between the ends, and a waist or neck
is formed before fracture.
Percentage reduction in area
Figure 4.2
(i) Between^ and B is the region in which Hooke’s
law applies and stress is directly proportion¬
al to strain. The gradient of AB is used when
determining Young’s modulus of elasticity (see
Chapter 3).
(ii) Point B is the limit of proportionality and
is the point at which stress is no longer
proportional to strain when a further load is
applied.
(iii) Point C is the elastic limit and a specimen
loaded to this point will effectively return to its
original length when the load is removed, i.e.
there is negligible permanent extension.
(iv) Point D is called the yield point and at this
point there is a sudden extension to J, with no
increase in load. The yield stress of the material
is given by:
yield stress =
load where yield begins to take place
original cross-sectional area
The yield stress gives an indication of the duc¬
tility of the material (see Chapter 3).
(original cross-sectional area) -
_ (final cross-sectional area) x 100°/
original cross-sectional area
The percentage reduction in area provides
information about the malleability of the
material (see Chapter 3). The value of stress at
point F is greater than at point E since although
the load on the specimen is decreasing as the
extension increases, the cross-sectional area is
also reducing.
(x) At point F the specimen fractures.
(xi) Distance GH is called the permanent elonga¬
tion and
permanent elongation
increase in length during test to destruction
original cross-sectional area
x 100 %
(xii) The point K is known as the upper yield point.
It occurs for constant load experiments, such
as when a hydraulic tensile testing machine
is used. It does not occur for constant strain
experiments, such as when a Hounsfield
tensometer is used.
Part Two
66 Mechanical Engineering Principles
4.2 Worked problems on
tensile testing
A tensile test is carried out on a mild
steel specimen. The results are shown in the
following table of values:
Load (kN)
0 10
23
32
Extension (mm)
0 0.023
0.053
0.074
Plot a graph of load against extension, and from the
graph determine (a) the load at an extension of
0.04 mm, and (b) the extension corresponding to a
load of 28 kN.
The load/extension graph is shown in Figure 4.3. From
the graph:
(a) when the extension is 0.04 mm, the load is 17.2 kN
(b) when the load is 28 kN, the extension is 0.065 mm
A tensile test is carried out on a mild
steel specimen of gauge length 40 mm and cross-
sectional area 100 mm 2 . The results obtained for
the specimen up to its yield point are given below:
Load (kN) 0 8 19 29 36
Extension 0 0.015 0.038 0.060 0.072
(mm)
The maximum load carried by the specimen is
50 kN and its length after fracture is 52 mm.
Determine (a) the modulus of elasticity, (b) the
ultimate tensile strength, (c) the percentage
elongation of the mild steel.
The load/extension graph is shown in Figure 4.4.
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
Extension/mm
Figure 4.4
(a) Gradient of straight line is given by:
BC 25000 N
(b)
A C 0.05 x 10 -3 m
= 500 x 10 6 N/m
, where
Young’s modulus of elasticity
f L '
= (gradient of graph) —
\ A J
L = 40 mm (gauge length) = 0.040 m and
area, A = 100 mm 2 = 100 x 10~ 6 m 2 .
Young’s modulus of elasticity
= (500 x 10 6 ) 0 040 = 200 x 10 9 Pa
100x10^
= 200 GPa
Ultimate tensile strength
maximum load
original cross-sectional area
50000N
100 x 10 -6 m 2
= 500 x 10 6 Pa = 500 MPa
Tensile testing 67
Percentage elongation
increase in length
original length
x 100
52-40
40
x 100
12
40
x 100-30%
The results of a tensile test are:
Diameter of specimen 15 mm; gauge length 40 mm;
load at limit of proportionality 85 kN; extension at
limit of proportionality 0.075 mm; maximum load
120 kN; final length at point of fracture 55 mm.
Determine (a) Young’s modulus of elasticity
(b) the ultimate tensile strength (c) the stress at
the limit of proportionality (d) the percentage
elongation.
(a) Young’s modulus of elasticity is given by:
F
E = stress _ A _ FL
strain x Ax
T
where the load at the limit of proportionality,
F= 85 kN- 85000 N,
L - gauge length - 40 mm = 0.040 m,
A = cross-sectional area
= = 7r ( Q - Q15 ) 2 - - 0.0001767 m 2 , and
4 4
x - extension - 0.075 mm = 0.000075 m.
Hence, Young’s modulus of elasticity,
_ FL _ (85000)(0.040)
~ Ax ~ (0.0001767)(0.000075)
= 256.6 x 10 9 Pa = 256.6 GPa
(b) Ultimate tensile strength
maximum load _ 120000
original cross-sectional area 0.0001767
= 679 x 10 6 Pa = 679 MPa
(c) Stress at limit of proportionality
load at limit of proportionality
cross-sectional area
85000
0.0001767
= 481 x 10 6 Pa = 481 MPa
(d) Percentage elongation
increase in length
original length
(55 - 40) mm
x 100
40 mm
x 100 = 37.5%
Now try the following Practice Exercise
Practice Exercise 28 Further problems on
tensile testing
1. What is a tensile test? Make a sketch of a
typical load/extension graph for a mild steel
specimen to the point of fracture and mark
on the sketch the following: (a) the limit of
proportionality (b) the elastic limit (c) the
yield point.
2. In a tensile test on a zinc specimen of gauge
length 100 mm and diameter 15 mm, a load of
100 kN produced an extension of 0.666 mm.
Determine (a) the stress induced (b) the
strain (c) Young’s modulus of elasticity.
[(a) 566 MPa (b) 0.00666 (c) 85 GPa]
3. The results of a tensile test are: Diameter of
specimen 20 mm, gauge length 50 mm, load
at limit of proportionality 80 kN, exten¬
sion at limit of proportionality 0.075 mm,
maximum load 100 kN, and final length
at point of fracture 60 mm. Determine
(a) Young’s modulus of elasticity (b) the
ultimate tensile strength (c) the stress at the
limit of proportionality (d) the percentage
elongation.
[(a) 169.8 GPa (b) 318.3 MPa'
(c) 254.6 MPa (d) 20%
4.3 Further worked problems
on tensile testing
A rectangular zinc specimen is sub¬
jected to a tensile test and the data from the test
is shown below. Width of specimen 40 mm;
breadth of specimen 2.5 mm; gauge length
120 mm.
Part Two
68 Mechanical Engineering Principles
Load (kN)
10 17 25 30 35 37.5 38.5 37 34 32
Extension (mm)
0.15 0.25 0.35 0.55 1.00 1.50 2.50 3.50 4.50 5.00
Fracture occurs when the extension is 5.0 mm and
the maximum load recorded is 38.5 kN. Plot the
load/extension graph and hence determine (a) the
stress at the limit of proportionality (b) Young’s
modulus of elasticity (c) the ultimate tensile
strength (d) the percentage elongation (e) the
stress at a strain of 0.01 (f) the extension at a stress
of 200 MPa.
A load/extension graph is shown in Figure 4.5.
Figure 4.5
(a) The limit of proportionality occurs at point P on
the graph, where the initial gradient of the graph
starts to change. This point has a load value of
26.5 kN.
Cross-sectional area of specimen
= 40 mm x 2.5 mm = 100 mm 2 = 100 x 10 6 m 2
Stress at the limit of proportionality is given by:
force 26.5 x 10 3 N
G — - = -
area 100xl0 _6 m 2
= 265 x 10 6 Pa = 265 MPa
(b) Gradient of straight line portion of graph is given
by:
BC
AC
25000N
0.35 x 10 _3 m
= 71.43 x 10 6 N/m
(c)
(d)
Young’s modulus of elasticity
= (gradient of graph)
V
= [71.43x10*
120x10
-3
v ' 100 xio -6
= 85.72 x 10 9 Pa = 85.72 GPa
Ultimate tensile strength
maximum load
original cross-sectional area
38.5 x 10 3 N
100 x 10 _6 m 2
= 385 x 10 6 Pa = 385 MPa
Percentage elongation
extension at fracture point
(e) Strain s =
original length
5.0mm 1AA . 1 _ n/
-x 100 = 4.17%
120 mm
extension x
x 100
original length /
extension x = el = 0.01 x 120
, from which,
= 1.20 mm.
From the graph, the load corresponding to an
extension of 1.20 mm is 36 kN.
Stress at a strain of 0.01 is given by:
force 36000 N
a = -=-
area 100xl0“ 6 m 2
= 360 x 10 6 Pa = 360 MPa
(f) When the stress is 200 MPa, then
force = area x stress
= (100 x 10~ 6 )(200 x 10 6 ) = 20 kN
From the graph, the corresponding extension is
0.30 mm
A mild steel specimen of cross-
sectional area 250 mm 2 and gauge length 100 mm
is subjected to a tensile test and the following data
is obtained: within the limit of proportionality, a
load of 75 kN produced an extension of 0.143 mm,
load at yield point = 80 kN, maximum load on
specimen = 120 kN, final cross-sectional area of
waist at fracture = 90 mm 2 , and the gauge length
had increased to 135 mm at fracture.
Determine for the specimen: (a) Young’s modulus
of elasticity (b) the yield stress (c) the tensile
Tensile testing 69
strength (d) the percentage elongation and (e) the
percentage reduction in area.
(a) Force F = 75 kN = 75000 N, gauge length
L = 100 mm = 0.1 m, cross-sectional area
A = 250 mm 2 = 250 x 10 -6 m 2 , and extension
x = 0.143 mm = 0.143 x 10 -3 m.
Young’s modulus of elasticity,
E _ stress _ F/A _ FL
strain x/L Ax
_ (75000)(0.1)
(250 x 10 _6 )(0.143 x 10 -3 )
= 210 x 10 9 Pa = 210 GPa
. load when yield begins to take place
(b) Yield stress =-—— £ --— : - - — £ -
original cross-sectional area
= 80000N
250 x 10 _6 m 2
= 320 x 10 6 Pa = 320 MPa
Tensile strength
maximum load
original cross- sectional area
__ 120000N
250 x 10 _6 m 2
= 480 x 10 6 Pa = 480 MPa
(d) Percentage elongation
increase in length during test to destruction
original length
= 135- 100 x 100 = 35%
100
(e) Percentage reduction in area
(original cross-sectional area)
-(final cross-sectional area)
original cross-sectional area
250-90
250
xl00 =
160
250
xlOO =64%
Now try the following Practice Exercise
Practice Exercise 29 Further problems on
tensile testing
1. A tensile test is carried out on a specimen
of mild steel of gauge length 40 mm and
diameter 7.42 mm. The results are:
At fracture the final length of the specimen
is 40.90 mm. Plot the load/extension graph
and determine (a) the modulus of elasticity
for mild steel (b) the stress at the limit
of proportionality (c) the ultimate tensile
strength (d) the percentage elongation.
(a) 210 GPa (b) 650 MPa'
(c) 890 MPa (d) 2.25%
2. An aluminium alloy specimen of gauge
length 75 mm and of diameter 11.28 mm was
subjected to a tensile test, with these results:
The specimen fractured at a load of 19.0 kN.
Determine (a) the modulus of elasticity of
the alloy (b) the percentage elongation.
[(a) 125 GPa (b) 0.413%]
3. An aluminium test piece 10 mm in diameter
and gauge length 50 mm gave the following
results when tested to destruction:
Load at yield point 4.0 kN, maximum load
6.3 kN, extension at yield point 0.036 mm,
diameter at fracture 7.7 mm.
Determine (a) the yield stress (b) Young’s
modulus of elasticity (c) the ultimate tensile
strength (d) the percentage reduction in area.
[(a) 50.93 MPa (b) 70.7 GPa'
(c) 80.2 MPa (d) 40.7%
4.4 Proof stress
Certain materials, such as high-tensile steel, aluminium
alloy, titanium, and so on, do not exhibit a definite yield
point in their stress-strain curves, unlike mild steel, as
shown in Figure 4.6.
Part Two
Part Two
70 Mechanical Engineering Principles
Figure 4.6 Stress-Strain curves for high tension steels,
aluminium alloys, etc.
In this case, the equivalent yield stresses are taken in
the form of either 0.1% or a 0.2% proof stress.
For 0.1% proof stress, a strain of e = 0.1% = y^ =
0.001, is set off on the horizontal strain axis of Figure 4.6
and a straight line is drawn parallel to the straight part of
the stress-strain curve from this point, near to the bottom
of the stress-strain curve where the angle is 6. The inter¬
section of this broken line with the stress-strain curve
gives the 0.1% proof stress, as shown in Figure 4.6.
0.2
For the 0.2% proof stress, a strain of s = 0.2% = y^ =
0.002, is set off on the horizontal strain axis of Figure
4.6 and a straight line is drawn from this point at an
angle 6 to the horizontal and parallel to the linear part
of the bottom of the stress-strain curve as shown in Fig¬
ure 4.6. The 0.2% proof stress is not twice the value of
the 0.1% proof stress! It may be about 10% larger.
Now try the following Practice Exercises
Practice Exercise 30 Short-answer questions
on tensile testing
1. What is a tensile test?
2. Which British Standard gives the standard
procedure for a tensile test?
3. With reference to a load/extension graph for
mild steel, state the meaning of (a) the limit
of proportionality (b) the elastic limit (c) the
yield point (d) the percentage elongation.
4. Define ultimate tensile strength.
5. Yield stress is the ratio
6. Define ‘percentage reduction in area’.
7. Briefly explain, with a diagram, what is
meant by ‘0.1% proof stress’.
Practice Exercise 31 Multiple-choice
questions on tensile
testing
(Answers on page 335)
A brass specimen having a cross-sectional area
of 100 mm 2 and gauge length 100 mm is sub¬
jected to a tensile test from which the following
information is obtained:
Load at yield point = 45 kN, maximum load
= 52.5 kN, final cross-sectional area of waist at
fracture = 75 mm 2 , and gauge length at fracture
= 110 mm.
For questions 1 to 4, select the correct answer
from the following list:
(a) 600 MPa (b) 525 MPa
(c) 33.33% (d) 10%
(e) 9.09% (f) 450 MPa
(g) 25% (h) 700 MPa
1. The yield stress.
2. The percentage elongation.
3. The percentage reduction in area.
4. The ultimate tensile strength.
References
Videos of the Tensile Test and Poisson’s Ratio experiment
can be obtained by visiting the website below:
www.routledge.com/cw/bird
For fully worked solutions to each of the problems in Practice Exercises 28 to 31 in this chapter,
go to the website:
www.routledge.com/cw/bird
Chapter 5
Why it is important to understand: Forces acting at a point
In this chapter the fundamental quantities, scalars and vectors, which form the very basis of all
branches of engineering, are introduced. A force is a vector quantity and, in this chapter, the resolution
of forces is introduced. Resolving forces is very important in structures, where the principle is used to
determine the strength of roof trusses, bridges, cranes, etc. Great lengths are gone to, to explain this
very fundamental skill, where step-by-step methods are adopted to help the reader understand this
very important procedure. The resolution of forces is also used in studying the motion of vehicles and
other particles in dynamics and in the case of the navigation of ships, aircraft, etc., the vectors take
the form of displacements, velocities and accelerations. This chapter gives a sound introduction to the
manipulation and use of scalars and vectors, by both graphical and analytical methods. The importance
of understanding the forces at a point is to appreciate the equilibrium of forces, whether they act on a
structure or a machine or a mechanism.
At the end of this chapter you should be able to:
• distinguish between scalar and vector quantities
• define ‘centre of gravity’ of an object
• define ‘equilibrium’ of an object
• understand the terms ‘coplanar’ and ‘concurrent’
• determine the resultant of two coplanar forces using (a) the triangle of forces method (b) the parallelogram of
forces method
• calculate the resultant of two coplanar forces using (a) the cosine and sine rules (b) resolution of forces
• determine the resultant of more than two coplanar forces using (a) the polygon of forces method (b) calcula¬
tion by resolution of forces
• determine unknown forces when three or more coplanar forces are in equilibrium
5.1 Scalar and vector quantities
Quantities used in engineering and science can be
divided into two groups:
(a) Scalar quantities have a size (or magnitude) only
and need no other information to specify them.
Thus, 10 centimetres, 50 seconds, 7 litres and
3 kilograms are all examples of scalar quantities.
(b) Vector quantities have both a size or magnitude
and a direction, called the line of action of the
quantity. Thus, a velocity of 50 kilometres per
hour due east, an acceleration of 9.81 metres per
second squared vertically downwards and a force
of 15 newtons at an angle of 30 degrees are all
examples of vector quantities.
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Two
72 Mechanical Engineering Principles
5.2 Centre of gravity and equilibrium
The centre of gravity of an object is a point where the
resultant gravitational force acting on the body may be
taken to act. For objects of uniform thickness lying in a
horizontal plane, the centre of gravity is vertically in line
with the point of balance of the object. For a thin uniform
rod the point of balance and hence the centre of gravity
is halfway along the rod as shown in Figure 5.1(a).
Figure 5.1
A thin flat sheet of a material of uniform thickness is
called a lamina and the centre of gravity of a rectangu¬
lar lamina lies at the point of intersection of its diago¬
nals, as shown in Figure 5.1(b). The centre of gravity of
a circular lamina is at the centre of the circle, as shown
in Figure 5.1(c).
An object is in equilibrium when the forces acting
on the object are such that there is no tendency for the
object to move. The state of equilibrium of an object
can be divided into three groups.
(i) If an object is in stable equilibrium and it is
slightly disturbed by pushing or pulling (i.e. a
disturbing force is applied), the centre of grav¬
ity is raised and when the disturbing force is re¬
moved, the object returns to its original position.
Thus a ball bearing in a hemispherical cup is in
stable equilibrium, as shown in Figure 5.2(a).
equilibrium
equilibrium
(c) Neutral
equilibrium
Figure 5.2
(ii) An object is in unstable equilibrium if, when a
disturbing force is applied, the centre of gravity
is lowered and the object moves away from its
original position. Thus, a ball bearing balanced
on top of a hemispherical cup is in unstable equi¬
librium, as shown in Figure 5.2(b).
(iii) When an object in neutral equilibrium has a
disturbing force applied, the centre of gravity re¬
mains at the same height and the object does not
move when the disturbing force is removed.
Thus, a ball bearing on a flat horizontal surface is
in neutral equilibrium, as shown in Figure 5.2(c).
5.3 Forces
When forces are all acting in the same plane, they are
called coplanar. When forces act at the same time and
at the same point, they are called concurrent forces.
Force is a vector quantity and thus has both a mag¬
nitude and a direction. A vector can be represented
graphically by a line drawn to scale in the direction of
the line of action of the force.
To distinguish between vector and scalar quantities,
various ways are used.
These include:
(i) bold print,
(ii) two capital letters with an arrow above them to
denote the sense of direction, for example, AB
where A is the starting point and B the end point
of the vector,
(iii) a line over the top of letters, for example, AB or a
(iv) letters with an arrow above, for example, a , A
(v) underlined letters, for example, a
(vi) xi + jy, where i and j are axes at right-angles to
each other; for example, 3/ + 4 j means 3 units in
the i direction and 4 units in the j direction, as
shown in Figure 5.3.
Figure 5.3
Forces acting at a point 73
(vii) a column matrix
/ \
a
~b
v /
; for example, the vector OA
shown in Figure 5.3 could be represented by
V
V 4 /
Thus, in Figure 5.3,
OA=OA = OA = 3/ + 4/ =
The method adopted in this text is to denote
vector quantities in bold print.
Thus, ab in Figure 5.4 represents a force of 5 newtons
acting in a direction due east.
3
4
and F 2 acting at point P as shown in Figure 5.6(a)
have exactly the same effect on point P as force
F shown in Figure 5.6(b), where F = F 2 ~F { and
acts in the direction of F 2 , since F 2 is greater than
F { . Thus F is the resultant of F { and F 2
P
Fi ^ -•-► F 2
(a)
Pm -► F
(f 2 -N
(b)
N
Scale
0 5 N
Tail ♦
a
■>- Nose
b
Figure 5.4
5.4 The resultant of two
coplanar forces
For two forces acting at a point, there are three pos¬
sibilities.
(a) For forces acting in the same direction and having
the same line of action, the single force having
the same effect as both of the forces, called the
resultant force or just the resultant, is the arith¬
metic sum of the separate forces. Forces of F x and
F 2 acting at point P, as shown in Figure 5.5(a),
have exactly the same effect on point P as force
F shown in Figure 5.5(b), where F = F { + F 2 and
acts in the same direction as F { and F 2 . Thus F is
the resultant of F { and F 2
P • > =— *~F 2
F 1
(a)
P •->■
(Fi + F 2 ) F
(b)
Figure 5.5
(b) For forces acting in opposite directions along the
same line of action, the resultant force is the arithme¬
tic difference between the two forces. Forces of F {
Figure 5.6
(c) When two forces do not have the same line of ac¬
tion, the magnitude and direction of the resultant
force may be found by a procedure called vector
addition of forces. There are two graphical meth¬
ods of performing vector addition, known as
the triangle of forces method (see Section 5.5)
and the parallelogram of forces method (see
Section 5.6).
Determine the resultant force of two
forces of 5 kN and 8 kN, (a) acting in the same
direction and having the same line of action
(b) acting in opposite directions but having the
same line of action.
(a) The vector diagram of the two forces acting in the
same direction is shown in Figure 5.7(a), which
assumes that the line of action is horizontal,
although since it is not specified, it could be in
any direction. From above, the resultant force F is
given by: F = F { + F 2 , i.e. F = (5 + 8) kN = 13 kN
in the direction of the original forces.
Scale
0 5 10 kN (force)
I _I_1_1_I_1_1_I_I_1_ I _1_I_L
5 kN
Pm ->->- 8 kN
(a)
5 kN
8 kN
—>►
Figure 5.7
Part Two
Part Two
74 Mechanical Engineering Principles
(b) The vector diagram of the two forces acting in op¬
posite directions is shown in Figure 5.7(b), again
assuming that the line of action is in a horizontal
direction. From above, the resultant force F is
given by: F = F 2 -F { , i.e. F= (8 - 5) kN = 3 kN
in the direction of the 8 kN force.
an angle of 60° from ab means 60° in an anti¬
clockwise direction.)
(iii) By measurement, the resultant ac is 30.5 units
long inclined at an angle of 35° to ab. That is,
the resultant force is 30.5 N, inclined at an an¬
gle of 35° to the 15 N force.
5.5 Triangle of forces method
A simple procedure for the triangle of forces method of
vector addition is as follows:
(i) Draw a vector representing one of the forces,
using an appropriate scale and in the direction
of its line of action.
(ii) From the nose of this vector and using the same
scale, draw a vector representing the second
force in the direction of its line of action.
(iii) The resultant vector is represented in both mag¬
nitude and direction by the vector drawn from
the tail of the first vector to the nose of the sec¬
ond vector.
Determine the magnitude and direc¬
tion of the resultant of a force of 15 N acting hori¬
zontally to the right and a force of 20 N, inclined at
an angle of 60° to the 15 N force. Use the triangle
of forces method.
Scale
0 5 10 15 20 N (force)
I_I_I_I_I
Figure 5.8
Using the procedure given above and with reference to
Figure 5.8:
(i) ab is drawn 15 units long horizontally.
(ii) From b, be is drawn 20 units long, inclined at an
angle of 60° to ab. (Note, in angular measure,
Find the magnitude and direction of
the two forces given, using the triangle of forces
method.
First force: 1.5 kN acting at an angle of 30°
Second force: 3.7 kN acting at an angle of-45°
From the above procedure and with reference to
Figure 5.9:
0 1 2 3 4 kN (force)
3.7kN
c
Figure 5.9
(i) ab is drawn at an angle of 30° and 1.5 units in
length.
(ii) From b , be is drawn at an angle of-45° and 3.7
units in length. (Note, an angle of-45° means
a clockwise rotation of 45° from a line drawn
horizontally to the right.)
(iii) By measurement, the resultant ac is 4.3 units
long at an angle of -25°. That is, the resultant
force is 4.3 kN at an angle of -25°
Now try the following Practice Exercise
Practice Exercise 32 Further problems on
the triangle of forces
method
1. Determine the magnitude and direction of
the resultant of the forces 1.3 kN and 2.7 kN,
having the same line of action and acting in
the same direction.
[4.0 kN in the direction of the forces]
Forces acting at a point 75
2. Determine the magnitude and direction of
the resultant of the forces 470 N and 538 N
having the same line of action but acting in
opposite directions.
[68 N in the direction of the 538 N force]
In questions 3 to 5, use the triangle of forces
method to determine the magnitude and direction
of the resultant of the forces given.
3. 13 N at 0° and 25 N at 30° [36.8 N at 20°]
4. 5 N at 60° and 8 N at 90° [12.6 N at 79°]
5. 1.3 kN at 45° and 2.8 kN at -30°
[3.4 kN at -8°]
5.6 The parallelogram of
forces method
A simple procedure for the parallelogram of forces
method of vector addition is as follows:
(i) Draw a vector representing one of the forces,
using an appropriate scale and in the direction
of its line of action.
(ii) From the tail of this vector and using the same
scale draw a vector representing the second force
in the direction of its line of action.
(iii) Complete the parallelogram using the two vec¬
tors drawn in (i) and (ii) as two sides of the paral¬
lelogram.
(iv) The resultant force is represented in both magni¬
tude and direction by the vector corresponding to
the diagonal of the parallelogram drawn from the
tail of the vectors in (i) and (ii).
Use the parallelogram of forces
method to find the magnitude and direction of the
resultant of a force of 250 N acting at an angle
of 135° and a force of 400 N acting at an angle
of-120°.
From the procedure given above and with reference to
Figure 5.10:
(i) ab is drawn at an angle of 135° and 250 units in
length
Scale
0 100 200 300 400 500 N (force)
b
Figure 5.10
(ii) ac is drawn at an angle of-120° and 400 units in
length
(iii) bd and cd are drawn to complete the parallelo¬
gram
(iv) ad is drawn. By measurement, ad is 413 units
long at an angle of-156°
That is, the resultant force is 413 N at an angle of -156°
Now try the following Practice Exercise
Practice Exercise 33 Further problems on
the parallelogram of
forces method
In questions 1 to 5, use the parallelogram of
forces method to determine the magnitude and
direction of the resultant of the forces given.
1. 1.7 N at 45° and 2.4 N at-60°
[2.6 N at -20°]
2. 9 Nat 126° and 14 Nat 223°
[15.7 Nat-172°]
3. 23.8 Nat -50° and 14.4 Nat 215°
[26.7 Nat -82°]
4. 0.7 kN at 147° and 1.3 kN at-71°
[0.86 kN at-101°]
5. 47 Nat 79° and 58 Nat 247°
[15.5 Nat-152°]
Part Two
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76 Mechanical Engineering Principles
5.7 Resultant of coplanar
forces by calculation
An alternative to the graphical methods of determining
the resultant of two coplanar forces is by calculation.
This can be achieved by trigonometry using the cosine
rule and the sine rule, as shown in Problem 5 following,
or by resolution of forces (see Section 5.10).
Use the cosine and sine rules to
determine the magnitude and direction of a force
of 8 kN acting at an angle of 50° to the horizontal
and a force of 5 kN acting at an angle of -30° to
the horizontal.
The space diagram is shown in Figure 5.11(a). A sketch
is made of the vector diagram, 0 a representing the
8 kN force in magnitude and direction and ab repre¬
senting the 5 kN force in magnitude and direction. The
resultant is given by length Ob. By the cosine rule,
a
(b) vector diagram
Now try the following Practice Exercise
Practice Exercise 34 Further problems
on the resultant of
coplanar forces by
calculation
1. Forces of 7.6 kN at 32° and 11.8 kN at 143°
act at a point. Use the cosine and sine rules
to calculate the magnitude and direction of
their resultant. [11.52 kN at 105°]
In questions 2 to 5, calculate the resultant of
the given forces by using the cosine and sine
rules.
2. 13 Nat 0° and 25 N at 30°
[36.84 Nat 19.83°]
3. 1.3 kN at 45° and 2.8 kN at-30°
[3.38 kN at-8.19°]
4. 9 Nat 126° and 14 Nat 223°
[15.69 N at- 171.67°]
5. 0.7 kN at 147° and 1.3 kN at-71°
[0.86 kN at- 100.94°]
5.8 Resultant of more than
two coplanar forces
Figure 5.11
0b 2 = 0 a 2 + ab 2 - 2(0 a)(ab) cosZOab
= 8 2 + 5 2 - 2(8)(5) cos 100°
(since ZOab = 180° - 50° - 30° = 100°)
= 64 + 25 - (-13.892) = 102.892
Hence, 0b = J\ 02.892 - 10.14 kN
5 10.14
By the sine rule, - - = - . --
smZaOb sin 100
5 sin 100°
from which, sin ZaOb = jq ^ = 0.4856
Hence ZaOb = sin -1 (0.4856) = 29.05°. Thus angle (p in
Figure 5.11(b) is 50° - 29.05° - 20.95°
Hence the resultant of the two forces is 10.14 kN
acting at an angle of 20.95° to the horizontal
For the three coplanar forces F 1? F 2 and acting at
a point as shown in Figure 5.12, the vector diagram
is drawn using the nose-to-tail method of Section 5.5.
The procedure is:
(i) Draw 0 a to scale to represent force F x in both
magnitude and direction (see Figure 5.13)
Forces acting at a point 77
b
Figure 5.13
(ii) From the nose of 0«, draw ab to represent
force F 2
(iii) From the nose of ab , draw be to represent
force F 3
(iv) The resultant vector is given by length 0 c in
Figure 5.13. The direction of resultant oc is from
where we started, i.e. point 0, to where we fin¬
ished, i.e. point c. When acting by itself, the re¬
sultant force, given by 0c, has the same effect
on the point as forces F { , F 2 and F 3 have when
acting together. The resulting vector diagram of
Figure 5.13 is called the polygon of forces.
Determine graphically the magnitude
and direction of the resultant of the following three
coplanar forces, which may be considered as acting
at a point:
Force A, 12 N acting horizontally to the right; force B ,
7 N inclined at 60° to forced; force C, 15 N inclined
at 150° to force A.
The space diagram is shown in Figure 5.14. The vector
diagram shown in Figure 5.15, is produced as follows:
F c = 15 N
Figure 5.14
c
(i) 0 a represents the 12 N force in magnitude and
direction
(ii) From the nose of 0 a, ab is drawn inclined at
60° to 0« and 7 units long
(iii) From the nose of ab , be is drawn 15 units long
inclined at 150° to 0 a (i.e. 150° to the horizontal)
(iv) 0c represents the resultant; by measurement,
the resultant is 13.8 N inclined at 0 = 80° to the
horizontal.
Thus the resultant of the three forces, F A9 F B and F c
is a force of 13.8 N at 80° to the horizontal.
The following coplanar forces are
acting at a point, the given angles being measured
from the horizontal: 100 N at 30°, 200 N at 80°,
40 N at -150°, 120 N at -100° and 70 N at -60°.
Determine graphically the magnitude and direction
of the resultant of the five forces.
The five forces are shown in the space diagram of
Figure 5.16. Since the 200 N and 120 N forces have
the same line of action but are in opposite sense, they
can be represented by a single force of 200 - 120, i.e.
80 N acting at 80° to the horizontal. Similarly, the
100 N and 40 N forces can be represented by a force
of 100 - 40, i.e. 60 N acting at 30° to the horizontal.
Hence the space diagram of Figure 5.16 may be rep¬
resented by the space diagram of Figure 5.17. Such a
simplification of the vectors is not essential but it is
easier to construct the vector diagram from a space
diagram having three forces, than one with five.
200 N
Figure 5.16
Figure 5.15
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78 Mechanical Engineering Principles
80N
Figure 5.17
The vector diagram is shown in Figure 5.18, 0 a repre¬
senting the 60 N force, ab representing the 8ON force
and be the 70 N force. The resultant, 0 c, is found by
measurement to represent a force of 112 N and angle
0 is 25°.
Figure 5.18
Thus, the five forces shown in Figure 5.16 may be
represented by a single force of 112 N at 25° to the
horizontal.
Now try the following Practice Exercise
Practice Exercise 35 Further problems on
the resultant of more
than two coplanar
forces
In questions 1 to 3, determine graphically the
magnitude and direction of the resultant of the
coplanar forces given which are acting at a point.
1. Force A, 12 N acting horizontally to the
right, force B , 20 N acting at 140° to force A ,
force C, 16 N acting 290° to force A.
[3.1 N at -45° to force A\
2. Force 1, 23 kN acting at 80° to the horizon¬
tal, force 2, 30 kN acting at 37° to force 1,
force 3, 15 kN acting at 70° to force 2.
'53.5 kN at 37° to force 1
(i.e. 117° to the horizontal)
3. Force P , 50 kN acting horizontally to the
right, force 0, 20 kN at 70° to force P, force
R , 40 kN at 170° to force P, force S , 80 kN at
300° to force P. [72 kN at -37° to force P]
4. Four horizontal wires are attached to a tele¬
phone pole and exert tensions of 30 N to the
south, 20 N to the east, 50 N to the north-east
and 40 N to the north-west. Determine the
resultant force on the pole and its direction.
[43.2 N at 39° east of north]
5.9 Coplanar forces in equilibrium
When three or more coplanar forces are acting at a
point and the vector diagram closes, there is no resul¬
tant. The forces acting at the point are in equilibrium.
Problem 8, A load of 200 N is lifted by two
ropes connected to the same point on the load,
making angles of 40° and 35° with the vertical.
Determine graphically the tensions in each rope
when the system is in equilibrium.
The space diagram is shown in Figure 5.19. Since
the system is in equilibrium, the vector diagram must
close. The vector diagram, shown in Figure 5.20, is
drawn as follows:
Figure 5.19
Forces acting at a point 79
Figure 5.20
(i) The load of 200 N is drawn vertically as shown
by 0 a
(ii) The direction only of force F x is known,
so from point a , ad is drawn at 40° to the verti¬
cal
(iii) The direction only of force F 2 is known,
so from point 0, 0 c is drawn at 35° to the
vertical
(iv) Lines ad and 0c cross at point b\ hence the
vector diagram is given by triangle 0 ab. By
measurement, ab is 119 N and 0 b is 133 N.
Thus the tensions in the ropes are F 1 = 119 N and
F 2 = 133 N
Five coplanar forces are acting on a
body and the body is in equilibrium. The forces
are: 12 kN acting horizontally to the right, 18 kN
acting at an angle of 75°, 7 kN acting at an angle
of 165°, 16 kN acting from the nose of the 7 kN
force, and 15 kN acting from the nose of the 16 kN
force. Determine the directions of the 16 kN and
15 kN forces relative to the 12 kN force.
With reference to Figure 5.21, 0 a is drawn 12 units
long horizontally to the right. From point a , ab is drawn
18 units long at an angle of 75°. From b, be is drawn
7 units long at an angle of 165°. The direction of the
16 kN force is not known, thus arc pq is drawn with
a compass, with centre at c, radius 16 units. Since the
forces are at equilibrium, the polygon of forces must
close. Using a compass with centre at 0, arc rs is drawn
having a radius 15 units. The point where the arcs
intersect is at d.
<i>
Figure 5.21
By measurement, angle 0 = 198° and a = 291°
Thus the 16 kN force acts at an angle of 198°
(or -162°) to the 12 kN force, and the 15 kN force
acts at an angle of 291° (or -69°) to the 12 kN force.
Now try the following Practice Exercise
Practice Exercise 36 Further problems on
coplanar forces in
equilibrium
1. A load of 12.5 N is lifted by two strings con¬
nected to the same point on the load, making
angles of 22° and 31° on opposite sides of
the vertical. Determine the tensions in the
strings. [5.86 N, 8.06 N]
2. A two-legged sling and hoist chain used for
lifting machine parts is shown in Figure 5.22.
Determine the forces in each leg of the sling
if parts exerting a downward force of 15 kN
are lifted. [9.96 kN, 7.77 kN]
Part Two
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80 Mechanical Engineering Principles
Figure 5.22
3. Four coplanar forces acting on a body are
such that it is in equilibrium. The vector
diagram for the forces is such that the 40 N
force acts vertically upwards, the 40 N force
acts at 65° to the 60 N force, the 100 N force
acts from the nose of the 40 N force and the
90 N force acts from the nose of the 100 N
force. Determine the direction of the 100 N
and 90 N forces relative to the 60 N force.
100 N force at 148° to the 60 N force,
90 N force at 277° to the 60 N force
5.10 Resolution of forces
A vector quantity may be expressed in terms of its
horizontal and vertical components. For example,
a vector representing a force of 10 N at an angle of
60° to the horizontal is shown in Figure 5.23. If the
horizontal line 0 a and the vertical line ab are construct¬
ed as shown, then 0 a is called the horizontal compo¬
nent of the 10 N force, and ab the vertical component
of the 10 N force.
b
By trigonometry, cos 60° = —, hence the horizontal
component, 0a = 10 cos 60°
Also, sin 60° = hence the vertical component,
ab = 10 sin 60°
This process is called finding the horizontal and ver¬
tical components of a vector or the resolution of a
vector, and can be used as an alternative to graphical
methods for calculating the resultant of two or more
coplanar forces acting at a point.
For example, to calculate the resultant of a 10N force
acting at 60° to the horizontal and a 20 N force acting at
-30° to the horizontal (see Figure 5.24) the procedure
is as follows:
Figure 5.24
(i) Determine the horizontal and vertical compo¬
nents of the 10N force, i.e. horizontal component,
0a = 10 cos 60° = 5.0 N, and vertical component,
ab= 10 sin 60° = 8.66 N
(ii) Determine the horizontal and vertical compo¬
nents of the 20 N force, i.e. horizontal compo¬
nent, 0 d = 20 cos(-30°) = 17.32 N, and vertical
component, cd = 20 sin(-30°) = -10.0 N
(iii) Determine the total horizontal component, i.e.
0a+ 0d=5.0 + 17.32 = 22.32 N
(iv) Determine the total vertical component, i.e.
ab + cd = 8.66 + (-10.0) =-1.34 N
(v) Sketch the total horizontal and vertical compo¬
nents as shown in Figure 5.25. The resultant of
the two components is given by length Or and, by
Pythagoras’ theorem,
Figure 5.23
Or = V22.32 2 + 1.34 2
= 22.36 N
Forces acting at a point 81
o
Total horizontal component = 22.32
Total vertical
component = -1.34
Figure 5.25
and using trigonometry,
1 34
angle 0 = tan ^ 1 ^32 =3 - 44 °
Hence the resultant of the 10 N and 20 N forces
shown in Figure 5.24 is 22.36 N at an angle
of -3.44° to the horizontal.
Forces of 5.0 N at 25° and 8.0 N at
112° act at a point. By resolving these forces into
horizontal and vertical components, determine their
resultant.
The space diagram is shown in Figure 5.26.
Figure 5.26
(i) The horizontal component of the 5.0 N force,
0 a = 5.0 cos 25° = 4.532, and the vertical
component of the 5.0 N force,
ab = 5.0 sin 25° = 2.113
(ii) The horizontal component of the 8.0 N force,
0c = 8.0 cos 112° =-2.991
The vertical component of the 8.0 N force,
cd = 8.0 sin 112° = 7.417
(iii) Total horizontal component
= 0a + 0c = 4.532 + (-2.997) = + 1.535
(iv) Total vertical component
- ab + cd = 2.113 + 7.417 - + 9.530
(v) The components are shown sketched in Figure
5.27.
Figure 5.27
By Pythagoras’ theorem, r=V1.535 2 + 9.530 2
= 9.653,
9.530
and by trigonometry, angle (p = tan 1 -
X • w/ m-S v/
= 80.85°
Hence the resultant of the two forces shown in
Figure 5.26 is a force of 9.653 N acting at 80.85° to
the horizontal.
Problems 9 and 10 demonstrate the use of resolution
of forces for calculating the resultant of two co-
planar forces acting at a point. However, the meth¬
od may be used for more than two forces acting at a
point, as shown in Problem 11.
Determine by resolution of forces
the resultant of the following three coplanar forces
acting at a point: 200 N acting at 20° to the hori¬
zontal; 400 N acting at 165° to the horizontal;
500 N acting at 250° to the horizontal.
A tabular approach using a calculator may be made as
shown below:
Horizontal component
Force 1 200 cos 20° = 187.94
Force 2 400 cos 165° = -386.37
Force 3 500 cos 250° =-171.01
Total horizontal component = -369.44
Vertical component
Force 1 200 sin 20° = 68.40
Force 2 400 sin 165°= 103.53
Part Two
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82 Mechanical Engineering Principles
Force 3 500 sin 250° = -469.85
Total vertical component = -297.92
The total horizontal and vertical components are shown
in Figure 5.28.
-369.44
Figure 5.28
Resultant r = V369.44 2 + 297.92 2
= 474.60, and
297 92
angle (j) = tan" 1 ^44 = 38.88°,
from which, a = 180° - 38.88° = 141.12°
Vertical components of the forces
150 sin 25° = 63.4
250 sin 90° - 250
200 sin 120° = 173.2
300 sin 200° = -102.6
60 sin 280° = -59.1
220 sin 320° = -141.4
Total vertical component, V — 183.5 N
H and V are shown plotted in Figure 5.29 and from
Pythagoras’ theorem:
Thus the resultant of the three forces given is
474.6 N acting at an angle of -141.12° (or +218.88°)
to the horizontal.
Determine by resolution the
magnitude and direction of the following six
concurrent coplanar forces: 150 N acting at 25°
to the horizontal, 250 N acting at 90° to the
horizontal, 200 N acting at 120° to the horizontal,
300 N acting at 200° to the horizontal, 60 N acting
at 280° to the horizontal, and 220 N acting at 320°
to the horizontal.
A tabular approach using a calculator may be made as
shown below:
Figure 5.29
R 2 = H 2 + V 2 and resultant,
R= Jh 2 + V 2 =a/(-67.1) 2 + (183.5) 2 = 195.4 N
= tan" 1
Angle d = tan 1
/ \
V
f \
183.5
v
67.1
7
= tan" 1 (2.7347) = 69.91°
Measuring from the positive horizontal axis, angle of
the resultant R,a= 180° - 69.91 ° = 110.09°
Thus the resultant of the six forces given is 195.4 N
acting at an angle of 110.09° to the horizontal.
Horizontal components of the forces
150 cos 25° = 135.9
250 cos 90° = 0
200 cos 120° = -100
300 cos 200°= -281.9
60 cos 280°= 10.42
220 cos 320°= 168.5
Total horizontal component, H— -67.1 N
Now try the following Practice Exercise
Practice Exercise 37 Further problems on
resolution of forces
1. Resolve a force of 23.0 N at an angle of 64°
into its horizontal and vertical components.
[10.08 N, 20.67 N]
2. Forces of 5 N at 21° and 9 N at 126° act
at a point. By resolving these forces into
Forces acting at a point 83
horizontal and vertical components, deter¬
mine their resultant. [9.09 N at 93.92°]
In questions 3 and 4, determine the magnitude
and direction of the resultant of the coplanar
forces given which are acting at a point, by reso¬
lution of forces.
3. Force A, 12 N acting horizontally to the
right, force B , 20 N acting at 140° to force A ,
force C, 16 N acting 290° to force A.
[3.06 N at -45.40° to force A]
4. Force 1, 23 kN acting at 80° to the horizon¬
tal, force 2, 30 kN acting at 37° to force 1,
force 3, 15 kN acting at 70° to force 2.
53.50 kN at 117.27°'
to the horizontal
5. Determine, by resolution of forces, the re¬
sultant of the following three coplanar forces
acting at a point: 10 kN acting at 32° to the
horizontal, 15 kN acting at 170° to the hori¬
zontal, 20 kN acting at 240° to the horizontal.
[18.82 kN at 210.03° to the horizontal]
6. The following coplanar forces act at a point:
force A, 15 N acting horizontally to the right,
force B , 23 N at 81 ° to the horizontal, force C,
7 N at 210° to the horizontal, force Z), 9 N at
265° to the horizontal, and force E , 28 N at
324° to the horizontal. Determine the resultant
of the five forces by resolution of the forces.
[34.96 N at -10.23° to the horizontal]
7. At a certain point, 12 different values of
coplanar and concurrent, radially outward
tensile forces are applied. The first force
is applied horizontally to the right and the
remaining 11 forces are applied at equally
spaced intervals of 30° anticlockwise. Start¬
ing from the first force and then 30° anti¬
clockwise to the second force, and so on, so
that the 12 forces encompass a complete cir¬
cle of 360°. The magnitude of the 12 forces,
in order of sequence, are: 30 kN, 250 kN,
200 kN, 180 kN, 160 kN, 140 kN, 120 kN,
100 kN, 80 kN, 60 kN, 40 kN and 20 kN.
Determine (a) the sum of the horizontal
components of the forces, //, (b) the sum
of the vertical components of the forces, V,
(c) the magnitude, R , and the direction of the
resultant force, 0 .
(a) H = -64.02 kN (b) V= 462.85 kN'
(c) R = 467.26 kN at 97.88°
anticlockwise from the horizontal
5.11 Summary
(a) To determine the resultant of two coplanar forces
acting at a point, four methods are commonly used.
They are:
by drawing:
(1) triangle of forces method, and
(2) parallelogram of forces method, and
by calculation:
(3) use of cosine and sine rules, and
(4) resolution of forces.
(b) To determine the resultant of more than two co¬
planar forces acting at a point, two methods are
commonly used. They are:
by drawing:
(1) polygon of forces method, and
by calculation:
(2) resolution of forces.
Now try the following Practice Exercises
Practice Exercise 38 Short-answer
questions on forces
acting at a point
1. Give one example of a scalar quantity and
one example of a vector quantity.
2. Explain the difference between a scalar and
a vector quantity.
3. What is meant by the centre of gravity of
an object?
4. Where is the centre of gravity of a rectan¬
gular lamina?
5. What is meant by neutral equilibrium?
6. State the meaning of the term ‘coplanar’.
7. What is a concurrent force?
8. State what is meant by a triangle of forces.
9. State what is meant by a parallelogram of
forces.
10. State what is meant by a polygon of forces.
Part Two
Part Two
84 Mechanical Engineering Principles
11. When a vector diagram is drawn represent¬
ing coplanar forces acting at a point, and
there is no resultant, the forces are in.
12. Two forces of 6 N and 9 N act horizontally
to the right. The resultant is .... N acting.
13. A force of 10 N acts at an angle of 50° and
another force of 20 N acts at an angle of
230°. The resultant is a force.N acting
at an angle of.... °.
14. What is meant by ‘resolution of forces’?
15. A coplanar force system comprises a 20 kN
force acting horizontally to the right, 30 kN
at 45°, 20 kN at 180° and 25 kN at 225°.
The resultant is a force of.N acting at
an angle of.... ° to the horizontal.
Practice Exercise 39 Multiple-choice
questions on forces
acting at a point
(Answers on page 335)
1. A physical quantity which has direction as
well as magnitude is known as a:
(a) force (b) vector
(c) scalar (d) weight.
2. Which of the following is not a scalar
quantity?
(a) velocity (b) potential energy
(c) work (d) kinetic energy.
3. Which of the following is not a vector
quantity?
(a) displacement
(b) density
(c) velocity
(d) acceleration.
4. Which of the following statements is false?
(a) Scalar quantities have size or magni¬
tude only
(b) Vector quantities have both magnitude
and direction
(c) Mass, length and time are all scalar
quantities
(d) Distance, velocity and acceleration are
all vector quantities.
5. If the centre of gravity of an object which
is slightly disturbed is raised and the object
returns to its original position when the dis¬
turbing force is removed, the object is said
to be in
(a) neutral equilibrium
(b) stable equilibrium
(c) static equilibrium
(d) unstable equilibrium.
6. Which of the following statements is false?
(a) The centre of gravity of a lamina is at
its point of balance.
(b) The centre of gravity of a circular lam¬
ina is at its centre.
(c) The centre of gravity of a rectangular
lamina is at the point of intersection of
its two sides.
(d) The centre of gravity of a thin uniform
rod is halfway along the rod.
7. The magnitude of the resultant of the
vectors shown in Figure 5.30 is:
(a) 2 N (b) 12 N
(c) 35 N (d) -2 N
5N
• -►
7N
Figure 5.30
8. The magnitude of the resultant of the vec¬
tors shown in Figure 5.31 is:
(a) 7 N (b) 5N
(c) IN (d) 12 N
4N A
->-3N
Figure 5.31
9. Which of the following statements is false?
(a) There is always a resultant vector required
to close a vector diagram representing
a system of coplanar forces acting at a
point, which are not in equilibrium.
(b) A vector quantity has both magnitude
and direction.
(c) A vector diagram representing a sys¬
tem of coplanar forces acting at a point
when in equilibrium does not close.
Forces acting at a point 85
(d) Concurrent forces are those which act
at the same time at the same point.
10. Which of the following statements is false?
(a) The resultant of coplanar forces of 1 N,
2 N and 3 N acting at a point can be 4 N.
(b) The resultant of forces of 6 N and
3 N acting in the same line of action
but opposite in sense is 3 N.
(c) The resultant of forces of 6 N and 3 N
acting in the same sense and having the
same line of action is 9 N.
(d) The resultant of coplanar forces of 4 N
at 0°, 3 N at 90° and 8 N at 180° is 15 N.
11. A space diagram of a force system is
shown in Figure 5.32. Which of the vector
diagrams in Figure 5.33 does not represent
this force system?
ION
Figure 5.32
30 N 30 N
10 N
Not to scale
Figure 5.33
12. With reference to Figure 5.34, which of the
following statements is false?
Figure 5.34
(a) The horizontal component of F A is
8.66 N
(b) The vertical component of F B is 10N
(c) The horizontal component of F c is 0
(d) The vertical component of F D is 4 N
13. The resultant of two forces of 3 N and 4 N
can never be equal to:
(a) 2.5 N (b) 4.5 N
(c) 6.5 N (d) 7.5 N
14. The magnitude of the resultant of the vec¬
tors shown in Figure 5.35 is:
(a) 5 N (b) 13 N
(c) 1 N (d) 63 N
Figure 5.35
7 N
For fully worked solutions to each of the problems in Practice Exercises 32 to 39 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Two
Chapter 6
Why it is important to understand: Simply supported beams
This chapter is very important for the design of beams which carry loads, acting transversely to them.
These structures are of importance in the design of buddings, bridges, cranes, ships, aircraft, automobiles,
and so on. The chapter commences with defining the moment of a force, and then, using equilibrium
considerations, demonstrates the principle of moments. This is a very important skill widely used in
designing structures. Several examples are given, with increasing complexity to help the reader acquire
these valuable skills. This chapter describes skids which are fundamental and extremely important in
many branches of engineering. The main reason though for understanding simply supported beams is
that such structures appear in the design of buddings, automobiles, aircraft, etc.
At the end of this chapter you should be able to:
• define a ‘moment’ of a force and state its unit
• calculate the moment of a force from M = Fx d
• understand the conditions for equilibrium of a beam
• state the principle of moments
• perform calculations involving the principle of moments
• recognise typical practical applications of simply supported beams with point loadings
• perform calculations on simply supported beams having point loads
• perform calculations on simply supported beams with couples
6.1 The moment of a force
When using a spanner to tighten a nut, a force tends
to turn the nut in a clockwise direction. This turning
effect of a force is called the moment of a force or,
more briefly, just a moment. The size of the moment
acting on the nut depends on two factors:
(a) the size of the force acting at right angles to the
shank of the spanner, and
(b) the perpendicular distance between the point of
application of the force and the centre of the nut.
In general, with reference to Figure 6.1, the moment M
of a force acting about a point P is: force x perpendicular
distance between the line of action of the force and P,
i.e. M-Fxd
Turning radius, d
Figure 6.1
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
Simply supported beams 87
The unit of a moment is the newton metre (N m).
Thus, if force F in Figure 6.1 is 7 N and distance d is
3 m, then the moment M is 7 N x 3 m, i.e. 21 N m.
A force of 15 N is applied to a
spanner at an effective length of 140 mm from
the centre of a nut. Calculate (a) the moment
of the force applied to the nut (b) the magnitude
of the force required to produce the same moment
if the effective length is reduced to 100 mm.
From above, M=Fxd , where M is the turning moment,
F is the force applied at right angles to the spanner and
d is the effective length between the force and the centre
of the nut. Thus, with reference to Figure 6.2(a):
P •-
\
15 N
140mm
A// = 2100 N mm
P% -
v
100mm
Figure 6.2
(a) Turning moment,
M- 15 Nx 140 mm = 2100 N mm
= 2100Nmmx—^— =2.1 Nm
1000 mm
(b) Turning moment, M is 2100 N mm and the
effective length d becomes 100 mm (see Figure
6.2(b)). Applying M-Fxd gives:
2100 N mm = F x 100 mm from which,
force,
2100 N mm
TOOmm
= 21 N
A moment of 25 N m is required
to operate a lifting jack. Determine the effective
length of the handle of the jack if the force applied
to it is: (a) 125 N (b) 0.4 kN
From above, moment M = F x d, where F is the force
applied at right angles to the handle and d is the effec¬
tive length of the handle. Thus:
(a)
(b)
25 N m = 125 N x d , from which effective length,
d - j ^ x 1000 mm = 200 mm
Turning moment M is 25 N m and the force F
becomes 0.4 kN, i.e. 400 N.
Since M = Fxd , then 25 N m = 400 N xd
Thus,
effective length, d -
25 N m _ 1
400 N “l6 m
x 1000 mm = 62.5 mm
16
Now try the following Practice Exercise
Practice Exercise 40 Further problems on
the moment of a force
1. Determine the moment of a force of 25 N
applied to a spanner at an effective length of
180 mm from the centre of a nut.
[4.5 Nm]
2. A moment of 7.5 N m is required to turn a
wheel. If a force of 37.5 N applied to the rim
of the wheel can just turn the wheel, calcu¬
late the effective distance from the rim to the
hub of the wheel. [200 mm]
3. Calculate the force required to produce
a moment of 27 N m on a shaft, when the
effective distance from the centre of the
shaft to the point of application of the force
is 180 mm. [150 N]
6.2 Equilibrium and the
principle of moments
If more than one force is acting on an object and the
forces do not act at a single point, then the turning
effect of the forces, that is, the moment of each of
the forces, must be considered.
Figure 6.3 shows a beam with its support (known as
its pivot or fulcrum) at P, acting vertically upwards,
and forces F { and F 9 acting vertically downwards at
distances a and b, respectively, from the fulcrum.
a
P
A
Figure 6.3
A beam is said to be in equilibrium when there is no
tendency for it to move. There are two conditions for
equilibrium:
(i) The sum of the forces acting vertically downwards
must be equal to the sum of the forces acting verti¬
cally upwards, i.e. for Figure 6.3, R p = F x + F 2
(ii) The total moment of the forces acting on a
beam must be zero; for the total moment to be
zero: the sum of the clockwise moments about
any point must be equal to the sum of the
Part Two
Part Two
88 Mechanical Engineering Principles
anticlockwise, or counter-clockwise, moments
about that point.
This statement is known as the principle of moments.
Hence, taking moments about P in Figure 6.3,
F 2 xb = the clockwise moment, and
F { x a = the anticlockwise, or counter¬
clockwise, moment
Thus for equilibrium: F 1 a = F 2 b
A system of forces is as shown in
Figure 6.4.
Fi p N ,7 N
I X 140 mm X
200 mm
d
Figure 6.4
(a) If the system is in equilibrium find the dis¬
tance d.
(b) If the point of application of the 5 N force is
moved to point P, distance 200 mm from the
support, and the 5 N force is replaced by an
unknown force F, find the value of F for the
system to be in equilibrium.
(a) From above, the clockwise moment M l is due to
a force of 7 N acting at a distance d from the sup¬
port; the support is called the fulcrum,
i.e. = 7 N x d
The anticlockwise moment M 2 is due to a force
of 5 N acting at a distance of 140 mm from the
fulcrum,
i.e. M 2 = 5 N x 140 mm
Applying the principle of moments, for the sys¬
tem to be in equilibrium about the fulcrum:
clockwise moment = anticlockwise moment
i.e. 7 N x d = 5 x 140 N mm
Hence, distance, d -
5 x HON mm
7N
= 100 mm
(b) When the 5 N force is replaced by force F at a dis¬
tance of 200 mm from the fulcrum, the new value
of the anticlockwise moment is F x 200. For the
system to be in equilibrium:
clockwise moment = anticlockwise moment
i.e. (7 xlOO) N mm = Fx 200 mm
Hence,
i r 700Nmm tcxt
new value of force, F -= 3.5 N
200 mm
A beam is supported on its fulcrum
at the point A, which is at mid-span, and forces act
as shown in Figure 6.5. Calculate (a) force F for
the beam to be in equilibrium (b) the new position
of the 23 N force when F is decreased to 21 N, if
equilibrium is to be maintained.
12 N F
I I
4
20 mm A d
v
■< —
■< -
->-
-<- >-
100 mm
80 mm
Figure 6.5
(a) The clockwise moment, is due to the 23 N
force acting at a distance of 100 mm from the
fulcrum, i.e. M\ = 23x 100 = 2300 N mm
There are two forces giving the anticlockwise
moment M 2 . One is the force F acting at a dis¬
tance of 20 mm from the fulcrum and the other a
force of 12 N acting at a distance of 80 mm.
Thus, M 2 = (F x 20) + (12 x 80) N mm
Applying the principle of moments about the ful¬
crum:
clockwise moment = anticlockwise moments
i.e. 2300 = (Fx 20) + (12x 80)
Hence F x 20 = 2300 - 960
i.e. force, F = = 67 N
(b) The clockwise moment is now due to a force
of 23 N acting at a distance of, say, d from the
fulcrum. Since the value of F is decreased to
21 N, the anticlockwise moment is
(21 x 20)+ (12x80) N mm.
Applying the principle of moments,
i.e.
23 xd
distance, d
(21 x 20)+ (12x 80)
420 + 960
23
1380
23
= 60 mm
For the centrally supported
uniform beam shown in Figure 6.6, determine the
values of forces F x and F 2 when the beam is in
equilibrium.
Simply supported beams 89
F i 3 m
7m F2
r ’
R
Figure 6.6
i
= 5kN
At equilibrium: (i) R = F { + F 2 i.e. 5 = F x + F 9 (1)
and (ii) Fj x 3 = F 2 x 7 (2)
From equation (1), F 2 = 5 -F {
Substituting for F ? in equation (2) gives:
Fj x 3 = (5 - F { ) x 7
i.e. 3Fj = 35 - 7F 1
10^ = 35
from which, F { - 3.5 kN
Since F 2 =5-F { , F 2 =1.5kN
Thus at equilibrium,
force Fj = 3.5 kN and force F 2 = 1.5 kN
Now try the following Practice Exercise
,20N , 30N
a y ye
R a>
k
20mm
50 mm
-< -—-
- ►
76mm
Figure 6.8
[R a = R b = 25 N]
4. The forces acting on a beam are as shown in
Figure 6.9. Neglecting the mass of the beam,
find the value of R 4 and distance d when the
beam is in equilibrium.
+
40 N
25 N
15 mm
t R a
60 N
-4-
-
d
35 mm
Figure 6.9
[5 N, 25 mm]
Practice Exercise 41 Further problems on
equilibrium and the
principle of moments
1 .
Determine distance d and the force
acting at the support A for the force system
shown in Figure 6.7, when the system is in
equilibrium.
1 kN
I A If
2.8 kN
R
A
140 mm d
Figure 6.7
[50 mm, 3.8 kN]
2. If the 1 kN force shown in Figure 6.7 is
replaced by a force F at a distance of
250 mm to the left of R 4 , find the value of F
for the system to be in equilibrium.
[560 N]
3. Determine the values of the forces acting
at A and B for the force system shown in
Figure 6.8.
6.3 Simply supported beams
having point loads
A simply supported beam is said to be one that rests on
two knife-edge supports and is free to move horizontally.
Two typical simply supported beams having loads
acting at given points on the beam, called point load¬
ing, are shown in Figure 6.10.
F
a
b
i
rc
A A
R a (a) R b
(b) R b
Figure 6.10
Part Two
Part Two
90 Mechanical Engineering Principles
A man whose mass exerts a force F vertically down¬
wards, standing on a wooden plank which is simply
supported at its ends, may, for example, be represented
by the beam diagram of Figure 6.10(a) if the mass of
the plank is neglected. The forces exerted by the sup¬
ports on the plank, R 4 and R B , act vertically upwards,
and are called reactions.
When the forces acting are all in one plane, the
algebraic sum of the moments can be taken about any
point.
For the beam in Figure 6.10(a) at equilibrium:
(i) r a + r b = f
and (ii) taking moments about A, F x a = R B (a + b)
(Alternatively, taking moments about C,
R A a = R B b )
For the beam in Figure 6.10(b), at equilibrium
(i) R a + R b = F { + F 2
and (ii) taking moments about B ,
R a (a + b) + F 2 c = F x b
Typical practical applications of simply supported
beams with point loadings include bridges, beams in
buildings, and beds of machine tools.
A beam is loaded as shown in Figure
6 . 11 .
, 2 kN , 7 kN , 3 kN
Y Y Y
A
A
-<—►0.2 m
/
_
0.5 m
0.8 m
^
1.0 m
(a)
| 2 kN | 7 kN |3 kN
R
A
(b)
k
Figure 6.11
Determine (a) the force acting on the beam support
at B (b) the force acting on the beam support at A,
neglecting the mass of the beam.
A beam supported as shown in Figure 6.11 is called a
simply supported beam.
clockwise moments = anticlockwise moments
(2 x 0.2) + (7 x 0.5) + (3 x 0.8) kNm = ^xl.0m
where R B is the force supporting the beam at B , as
shown in Figure 6.11(b).
Thus
i.e.
(0.4 + 3.5 + 2.4) kN m = R B x 1.0 m
D _ 6.3 kNm
Rb ~ 1.0m
6.3 kN
(b) For the beam to be in equilibrium, the forces
acting upwards must be equal to the forces acting
downwards, thus
R a + R b = (2 + 7 + 3) kN = 12 kN
R b = 6.3 kN,
thus R a - 12 ~R b - 12 - 6.3
= 5.7 kN
For the beam shown in Figure
6.12 calculate (a) the force acting on support^
(b) distance d, neglecting any forces arising from
the mass of the beam.
Figure 6.12
(a) From Section 6.2,
(the forces acting in an upward direction) =
(the forces acting in a downward direction)
Hence (R A + 40) N = (10 + 15 + 30) N
R a = 10+ 15 + 30-40 = 15N
(b) Taking moments about the left-hand end of the
beam and applying the principle of moments
gives:
clockwise moments = anticlockwise moments
(10 x 0.5) + (15 x 2.0) N m + 30 N x d
- (15 x 1.0) + (40 x 2.5) N m
i.e. 35Nm + 30Nxd= 115Nm
(115 - 35) Nm
from which, distance, d =--
(a) Taking moments about point A and applying the
principle of moments gives:
= 2.67 m
Simply supported beams 91
A metal bar AB is 4.0 m long and
is simply supported at each end in a horizontal
position. It carries loads of 2.5 kN and 5.5 kN at
distances of 2.0 m and 3.0 m, respectively, from^f.
Neglecting the mass of the beam, determine the
reactions of the supports when the beam is in
equilibrium.
The beam and its loads are shown in Figure 6.13.
2.5 kN 5.5 kN
2.0 m
1.0 m
^^
^^
A I-IB
A A
Figure 6.13
At equilibrium, R 4 + R B = 2.5 + 5.5 = 8.0 kN (1)
Taking moments about A,
clockwise moments = anticlockwise moments
i.e. (2.5 x 2.0) + (5.5 x 3.0) = 4.0 R B
or 5.0 + 16.5 = 4.0 R B
from which, R B = = 5.375 kN
From equation (1), R 4 = 8.0 - 5.375 = 2.625 kN
Thus the reactions at the supports at equilibrium
are 2.625 kN at^4 and 5.375 kN atB
Abeam PQ is 5.0 m long and is simply
supported at its ends in a horizontal position as shown
in Figure 6.14. Its mass is equivalent to a force of
400 N acting at its centre as shown. Point loads of
12 kN and 20 kN act on the beam in the positions
shown. When the beam is in equilibrium, determine
(a) the reactions of the supports, R p and Rq, and
(b) the position to which the 12 kN load must be
moved for the force on the supports to be equal.
12 kN
400 N 20 kN
^
^- )
1.2 m >
f 1.3 m
t t
1.5 m
Q
Rc
R
Q
Figure 6.14
(a) At equilibrium,
R p + R Q = 12+ 0.4+ 20 = 32.4 kN (1)
Taking moments about P:
clockwise moments = anticlockwise moments
i.e. (12 x 1.2) +(0.4 x 2.5)
+ (20 x 3.5) = (Rq x 5.0)
14.4 + 1.0 + 70.0 = 5.0 Rq
85.4
from which, Rq = -yy = 17.08 kN
From equation (1), R p = 32.4 -Rq
= 32.4-17.08
= 15.32 kN
(b) For the reactions of the supports to be equal,
32 4
R p — R q — -y- = 16.2 kN
Let the 12 kN load be at a distance d metres from
P (instead of at 1.2 m from P). Taking moments
about point P gives:
(12 x d) + (0.4 x 2.5) + (20 x 3.5) = 5.0 R ()
i.e. 12 d+ 1.0+ 70.0 = 5.0 x 16.2
and \2d =81.0-71.0
from which, d - = 0.833 m
Hence the 12 kN load needs to be moved to a
position 833 mm from P for the reactions of the
supports to be equal (i.e. 367 mm to the left of its
original position).
A uniform steel girder AB is 6.0 m
long and has a mass equivalent to 4.0 kN acting at
its centre. The girder rests on two supports at C and
B as shown in Figure 6.15. A point load of 20.0 kN
is attached to the beam as shown. Determine the
value of the force F that causes the beam to just lift
off the support B.
3.0 m
4.0 kN
H
2.5 m
20.0 kN
Figure 6.15
Part Two
Part Two
92 Mechanical Engineering Principles
At equilibrium, R c + R B - F + 4.0 + 20.0.
When the beam is just lifting off of the support B , then
R b = 0, hence R c = (F + 24.0)kN.
Taking moments about A:
Clockwise moments = anticlockwise moments
i.e. (4.0 x 3.0) + (5.0 x 20.0) = (R c x 2.5) + (R B x 6.0)
i.e. 12.0 + 100.0 = (F + 24.0) x 2.5 + (0)
i.e. T^=OF + 24.0)
from which, F - 44.8 - 24.0 = 20.8 kN
3. For the force system shown in Figure 6.18,
determine distance d for the forces R A
and R b to be equal, assuming equilibrium
conditions. [80 m]
10 N
I
15 N
i
of
i
i
*
^Um
20 m
20 m
25 N
\
R
B
Figure 6.18
i.e. the value of force F which causes the beam to
just lift off the support B is 20.8 kN
Now try the following Practice Exercise
Practice Exercise 42 Further problems on
simply supported
beams having point
loads
1. Calculate the force R 4 and distance d for the
beam shown in Figure 6.16. The mass of the
beam should be neglected and equilibrium
conditions assumed.
0.2 kN
2.7 kN
T '
t
| Ra
| 1.3 kN
12 mm 10 mm
d
15 mm
Figure 6.16
[2.0 kN, 24 mm]
2. For the force system shown in Figure 6.17,
find the values of F and d for the system to
be in equilibrium.
1.4 kN
0.7 kN
F
<->
14 mm
2.3 kN
12 mm
cf
5 mm
0.8 kN
Figure 6.17
10 kN 16 kN
4
I
A
j
k |
2m*
< 2 m
2 m ^
^ 2m^l
B
R-
R'
Figure 6.19
4. A simply supported beam AB is loaded as
shown in Figure 6.19. Determine the load F
in order that the reaction at A is zero.
[36 kN]
5. A uniform wooden beam, 4.8 m long, is
supported at its left-hand end and also at
3.2 m from the left-hand end. The mass of
the beam is equivalent to 200 N acting verti¬
cally downwards at its centre. Determine the
reactions at the supports. [50 N, 150 N]
6. For the simply supported beam PQ shown in
Figure 6.20, determine (a) the reaction at each
support (b) the maximum force which can be
applied at Q without losing equilibrium.
4 kN 6 kN 5 kN
1.5 m
4.0 m
1.5 m
2.0 m
f i
f T
Ri R 2
Figure 6.20
[(a) Rj = 3 kN, R 2 = 12 kN (b) 15.5 kN]
7. A uniform beam AB is 12.0 m long and is sup¬
ported at distances of 2.0 m and 9.0 m from A.
Loads of 60 kN, 104 kN, 50 kN and 40 kN
act vertically downwards at^4, 5.0 m from A,
7.0 m from A and at B. Neglecting the mass
[1.0 kN, 64 mm]
Simply supported beams 93
of the beam, determine the reactions at the
supports. [133.7 kN, 120.3 kN]
8. A uniform girder carrying point loads is
shown in Figure 6.21. Determine the value
of load F which causes the beam to just lift
off the support B.
F 10 kN 4 kN 5 kN
2 m
r 4 m
. 3m,
2 m,
Y >
f
A
t
4 m 1
Figure 6.21
[3.25 kN]
6.4 Simply supported beams with
couples
The procedure adopted here is a simple extension to
Section 6.3, but it must be remembered that the units of
a couple are in: N m, N mm, kN m, etc., unlike that of
a force. The method of calculating reactions on beams
due to couples will now be explained with the aid of
worked problems.
Determine the end reactions for
the simply supported beam of Figure 6.22, which
is subjected to an anticlockwise couple of 5 kN m
applied at mid-span.
5 kN m
Figure 6.22
i.e. R a
x 3 m = 5 kN m
(6.1)
from which,
*4=fkN
or
R a = 1.667 kN
(6.2)
Resolving forces vertically gives:
Upward forces = downward forces
i.e. R a +R b = 0 (6.3)
It should be noted that in equation (6.3) the 5 kN m
couple does not appear; this is because it is a couple
and not a force.
From equations (6.2) and (6.3), R B --R 4 - -1.667 kN
i.e. R b acts in the opposite direction to R A , so that
R b and R a also form a couple that resists the 5 kN m
couple.
Determine the end reactions for the
simply supported beam of Figure 6.23, which is
subjected to an anticlockwise couple of 5 kN m at
the point C.
^_5 kN m
■< -
■< -2m-
C A
-
Figure 6.23
Taking moments about B gives:
^x3m = 5kNm (6.4)
from which, R A = ^ kN
or R a = 1.667 kN
Resolving forces vertically gives:
i.e. R 4 +R b = 0
from which, R B = -R 4 = -1.667 kN
Taking moments about B:
Now the reaction R A exerts a clockwise moment about
B given by: R 4 x 3 m
Additionally, the couple of 5 kN m is anticlockwise
and its moment is 5 kN m regardless of where it is
placed.
Clockwise moments about B =
anticlockwise moments about B
It should be noted that the answers for the reactions
are the same for Problems 11 and 12, thereby proving
by induction that the position of a couple on a beam,
simply supported at its ends, does not affect the values
of the reactions.
Determine the reactions for the
simply supported beam of Figure 6.24.
Part Two
Part Two
94 Mechanical Engineering Principles
10kNm 8kNm 6kNm
A
R.
-<—
—► ■< —
k
C
w
D
E a
w
1 m
,1m.
1 m
“-“
,1m.
^
^
B
x
R
B
Figure 6.24
Taking moments about B gives:
R a x 4 m + 8 kN m = 10 kN m + 6 kN m
= 10 + 6-8 = 8
L A
8
i.e.
from which,
Resolving forces vertically gives:
*A= 4= 2kN
from which,
Ra + Rfi — 0
R B = -R A = -2kN
Determine the reactions for the
simply supported beam of Figure 6.25.
10 kN m 8 kN m 6 kN
A
R
A
-< —
~~ t
A C
—>► -<—
D E a
1m .
1 m
1m | 1m
B
R
B
Figure 6.25
Taking moments about B gives:
i.e.
R a x 4 m + 8 kN m + 6 kN x 1 m = 10 kN m
4 R a = 10 - 8 - 6 = - 4
from which, R A = - — = -1 kN (acting downwards)
Resolving forces vertically gives:
R ^ Rb 6 = 0
from which, R B - -R A - 6 = - (-1) - 6
i.e. R b = 1 - 6 = -5 kN (acting downwards)
Now try the following Practice Exercises
Practice Exercise 43 Further problems on
simply supported
beams with couples
For each of the following problems, determine the
reactions acting on the simply supported beams:
1.
5 kN m
4
R.
A
-<—
—3 m—
—>►
A
-<—2 m—
B
R
B
Figure 6.26
[R a = - 1 kN, R r = 1 kN]
B
2 .
5 kN m
4
R,
a
A
2.5 m
2.5 m
* X ^
B
R
B
Figure 6.27
[R a = - 1 kN, R r = 1 kN]
B
3.
10 kN m 6 kN m 12 kN m
R
A
—
1 1
-<—
2 m
2 m
—►
2 m
>k
—►
2m
R
B
Figure 6.28
[R a = 1 kN, R r = - 1 kN]
B
4.
10 kN m
10 kN m
1 m
■<—►
R
A
5 m
2 m
<-*•
Figure 6.29
\r a = q,r b = 0]
Simply supported beams 95
5.
10 kNm
1 m 3m
<—►<-
10 kNm
-<—
4 m
- >
4. State the principle of moments.
5. What is meant by a simply supported beam?
6. State two practical applications of simply
supported beams.
7. Why does a couple not have a vertical com¬
ponent of force?
Figure 6.30
[*, = 0 ,^= 0 ]
8 kN
10 kN m
-< —
Y
A
2 m , 2 m
-<-- — -►
A
—►2 m
■< - ►
Figure 6.31
[R a = 7 kN, R b = 1 kN]
7.
Practice Exercise 45 Multiple-choice
questions on simply
supported beams
(Answers on page 335)
1. A force of 10 N is applied at right angles
to the handle of a spanner, 0.5 m from the
centre of a nut. The moment on the nut is:
(a) 5 N m (b) 2 N/m
(c) 0.5 m/N (d) 15 Nm
2. The distance d in Figure 6.33 when the
beam is in equilibrium is:
(a) 0.5 m (b) 1.0 m
(c) 4.0 m (d) 15 m
10 kNm 12 kNm
t
2 m
k
r a
2 m
i
2 m
k
r b
2 m
1 m
Figure 6.32
Figure 6.33
[R a = - 333 N, R b = 333 N]
Practice Exercise 44 Short-answer
questions on simply
supported beams
1. The moment of a force is the product of.
and.
2. When a beam has no tendency to move it is
in.
3. With reference to Figure 6.34, the clock¬
wise moment about A is:
(a) 70 Nm (b) 10 Nm
(c) 60 N m (d) 5 x R B N m
>
10N
20 N
f
A
Ra
1 m
-<—
3 m
_ ^
A
- >-
5 m
-< -
3. State the two conditions for equilibrium of a
beam.
Figure 6.34
Part Two
Part Two
96 Mechanical Engineering Principles
4. The force acting at B (i.e. R B ) in Figure
6.34 is:
(a) 16 N (b) 20 N
(c) 5 N (d) 14 N
5. The force acting at A (i.e. R 4 ) in Figure
6.34 is:
(a) 16 N (b) ION
(c) 15 N (d) 14 N
6. Which of the following statements is false
for the beam shown in Figure 6.35 if the
beam is in equilibrium?
(a) The anticlockwise moment is 27 N
(b) The force F is 9 N
(c) The reaction at the support R is 18 N
(d) The beam cannot be in equilibrium for
the given conditions
E
6N 3N q F
3.0 m
3.0 m
i_i_i
A
R
Figure 6.35
7. With reference to Figure 6.36, the reaction
r a is:
(a) ION (b) 30 N
(c) 20 N (d) 40 N
ION 20 N 10N
1 0.5m
1 ^ w
0.5 m
0.5m
0.5m
r
^^
>
r
>
t t
Ra Rb
Figure 6.36
8. With reference to Figure 6.36, when mo¬
ments are taken about R 4 , the sum of the
anticlockwise moments is:
(a) 25 Nm (b) 20 N m
(c) 35 Nm (d) 30 N m
9. With reference to Figure 6.36, when mo¬
ments are taken about the right-hand end,
the sum of the clockwise moments is:
(a) 10 Nm (b) 20 N m
(c) 30 N m (d) 40 N m
10. With reference to Figure 6.36, which of the
following statements is false?
(a) (5 + R b ) = 25 Nm
(b) R a = R b
(c) (10 x 0.5) = (10 x 1) + (10 x 1.5) + R a
(d) * 1 +* 2( = 40N
11. A beam simply supported at its ends is
subjected to two intermediate couples
of 4 kN m clockwise and 4 kN m anti¬
clockwise. The values of the end reactions
are:
(a) 4 kN (b) 8 kN
(c) zero (d) unknown
For fully worked solutions to each of the problems in Practice Exercises 40 to 45 in this chapter,
go to the website:
www.routledge.com/cw/bird
Revision Test 3 Forces, tensile testing and beams
This Revision Test covers the material contained in Chapters 3 to 6. The marks for each question are shown in
brackets at the end of each question.
1. A metal bar having a cross-sectional area of
80 mm 2 has a tensile force of 20 kN applied to it.
Determine the stress in the bar. (4)
2. (a) A rectangular metal bar has a width of 16 mm
and can support a maximum compressive stress
of 15 MPa; determine the minimum breadth of
the bar when loaded with a force of 6 kN.
(b) If the bar in (a) is 1.5 m long and decreases
in length by 0.18 mm when the force is ap¬
plied, determine the strain and the percentage
strain. (7)
3. A wire is stretched 2.50 mm by a force of 400 N.
Determine the force that would stretch the wire
3.50 mm, assuming that the elastic limit is not
exceeded. (5)
4. A copper tube has an internal diameter of 140 mm
and an outside diameter of 180 mm and is used
to support a load of 4 kN. The tube is 600 mm
long before the load is applied. Determine, in
micrometres, by how much the tube contracts
when loaded, taking the modulus of elasticity for
copper as 96 GPa. (8)
5. The results of a tensile test are: diameter of speci¬
men 21.7 mm; gauge length 60 mm; load at limit
of proportionality 50 kN; extension at limit of pro¬
portionality 0.090 mm; maximum load 100 kN;
final length at point of fracture 75 mm.
Determine (a) Young’s modulus of elasticity
(b) the ultimate tensile strength (c) the stress at
the limit of proportionality (d) the percentage
elongation. (10)
6. A force of 25 N acts horizontally to the right and a
force of 15 N is inclined at an angle of 30° to the
25 N force. Determine the magnitude and direc¬
tion of the resultant of the two forces using (a) the
triangle of forces method (b) the parallelogram of
forces method and (c) by calculation. (12)
7. Determine graphically the magnitude and direc¬
tion of the resultant of the following three co-
planar forces, which may be considered as acting
at a point.
Force P, 15 N acting horizontally to the right
Force Q, 8 N inclined at 45° to force P
Force R , 20 N inclined at 120° to force P. (7)
8. Determine by resolution of forces the resultant
of the following three coplanar forces acting at
a point: 120 N acting at 40° to the horizontal;
250 N acting at 145° to the horizontal; 300 N
acting at 260° to the horizontal. (8)
9. A moment of 18 N m is required to operate a lift¬
ing jack. Determine the effective length of the
handle of the jack (in millimetres) if the force
applied to it is (a) 90 N (b) 0.36 kN. (6)
10. For the centrally supported uniform beam shown
in Figure RT3.1, determine the values of forces
F { and P 2 when the beam is in equilibrium. (7)
Fl 2m
5 m F >
T
_f
R = 5
.6 kN
Figure RT3.1
11. For the beam shown in Figure RT3.2 calculate
(a) the force acting on support 0, (b) distance d ,
neglecting any forces arising from the mass of the
beam. (7)
5 N
T ^
20 N
f y
10N
f
P Q
|1m A
m R p = 15
- ►
6m
N ^
R o
^
9m
^_w.
d
Figure RT3.2
12. A beam of length 3 m is simply supported at its
ends. If a clockwise couple of 4 kN m is placed
at a distance of 1 m from the left hand support,
determine the end reactions. (4)
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 3,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
Part Two
Chapter 7
Forces in structures
Why it is important to understand: Forces in structures
In this chapter it will be shown how the principles described in Chapter 5 on forces acting at a point
can be used to determine the internal forces in the members of a truss, due to externally applied loads.
In countries where there is a lot of rain, such structures are used to support the sloping roofs of the
building; the externally applied loads acting on the pin-jointed trusses are usually due to snow and
self-weight, and also due to wind. Tie bars and struts are defined, Bow’s notation is explained and
internal forces in a truss are calculated by different methods. The importance of understanding forces
in structures is that these appear in the design of buddings, ships, automobiles etc.
At the end of this chapter you should be able to:
• recognise a pin-jointed truss
• recognise a mechanism
• define a tie bar
• define a strut
• understand Bow’s notation
• calculate the internal forces in a truss by a graphical method
• calculate the internal forces in a truss by the ‘method of joints’
• calculate the internal forces in a truss by the ‘method of sections’
7.1 Introduction
As stated above, in this chapter it will be shown how the
principles described in Chapter 5 can be used to deter¬
mine the internal forces in the members of a truss, due
to externally applied loads. The definition of a truss
is that it is a frame where the joints are assumed to be
frictionless and pin-jointed, and that all external loads
are applied to the pin joints. In countries where there
is a lot of rain, such structures are used to support the
sloping roofs of the building, as shown in Figure 7.1.
The externally applied loads acting on the pin-jointed
trusses are usually due to snow and self-weight, and also
due to wind, as shown in Figure 7.1. In Figure 7.1, the
snow and self-weight loads act vertically downwards and
the wind loads are usually assumed to act horizontally.
Thus, for structures such as that shown in Figure 7.1,
where the externally applied loads are assumed to act at the
pin-joints, the internal members of the framework resist
the externally applied loads in tension or compression.
Figure 7.1 Pin-jointed truss
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
Forces in structures 99
Members of the framework that resist the externally ap¬
plied loads in tension are called ties and members of the
framework which resist the externally applied loads in
compression are called struts, as shown in Figure 7.2.
the two bottom joints near the mid-length of the bottom
horizontal.
Three methods of analysis will be used in this
chapter - one graphical and two analytical methods.
G =3 —- —► G 9 —
(a) Tie (b) Strut
Figure 7.2 Ties and stmts with external loads
The internal resisting forces in the ties and struts will
act in the opposite direction to the externally applied
loads, as shown in Figure 7.3.
G—C>—-<1-^ 3 G <1 t> =0
(a) Tie in tension (b) Strut in compression
Figure 7.3 Internal resisting forces in ties and stmts
The methods of analysis used in this chapter break¬
down if the joints are rigid (welded), or if the loads
are applied between the joints. In these cases, flexure
occurs in the members of the framework, and other
methods of analysis have to be used, as described in
Chapters 6 and 8. It must be remembered, however,
that even if the joints of the framework are smooth
and pin-jointed and also if externally applied loads are
placed at the pin-joints, members of the truss in com¬
pression can fail through structural failure (see refer¬
ences [1] and [2] on page 111).
It must also be remembered that the methods used
here cannot be used to determine forces in statically
indeterminate pin-jointed trusses, nor can they be
used to determine forces in mechanisms. Statically
indeterminate structures are so called because they
cannot be analysed by the principles of statics alone.
Typical mechanisms are shown in Figure 7.4; these are
not classified as structures because they are not firm
and can be moved easily under external loads.
To make the mechanism of Figure 7.4(a) into a simple
statically determinate structure, it is necessary to add
one diagonal member joined to a top joint and an ‘op¬
posite’ bottom joint. To make the mechanism of Figure
7.4(b) into a statically determinate structure, it is nec¬
essary to add two members from the top joint to each of
7.2 Worked problems on mechanisms
and pin-jointed trusses
Show how the mechanism of Figure
7.4(a) can be made into a statically determinate
structure.
The two solutions are shown by the broken lines of
Figures 7.5(a) and (b), which represent the placement
of additional members.
Figure 7.5
Show how the mechanism of Figure
7.4(b) can be made into a statically determinate
truss.
The solution is shown in Figure 7.6, where the broken
lines represent the placement of two additional members.
Figure 7.6
Show how the mechanism of Figure
7.4(a) can be made into a statically indeterminate
truss.
The solution is shown in Figure 7.7, where the broken
lines represent the addition of two members, which are
not joined where they cross.
Figure 7.7
Part Two
Part Two
100 Mechanical Engineering Principles
Why is the structure of Figure 7.7
said to be statically indeterminate?
As you can only resolve vertically and horizontally at
the joints A and B , you can only obtain four simultane¬
ous equations. However, as there are five members, each
with an unknown force, you have one unknown force
too many. Thus, using the principles of statics alone, the
structure cannot be satisfactorily analysed; such struc¬
tures are therefore said to be statically indeterminate.
7.3 Graphical method
In this case, the method described in Chapter 5 will be
used to analyse statically determinate plane pin-jointed
trusses. The method will be described with the aid of
worked examples.
Determine the internal forces that
occur in the plane pin-jointed truss of Figure 7.8,
due to the externally applied vertical load of 3 kN.
3 kN
I
Figure 7.8
Firstly, we will fill the spaces between the forces with
upper case letters of the alphabet, as shown in Figure
7.8. It should be noted that the only reactions are the
vertical reactions R { and Ry this is because the only
externally applied load is the vertical load of 3 kN, and
there is no external horizontal load. The capital let¬
ters A, B, C and D can be used to represent the forces
between them, providing they are taken in a clockwise
direction about each joint. Thus the letters AB represent
the vertical load of 3 kN. Now as this load acts vertically
downwards, it can be represented by a vector ab , where
the magnitude of ab is 3 kN and it points in the direction
from a to b. As ab is a vector, it will have a direction
as well as a magnitude. Thus ab will point downwards
from a to b as the 3 kN load acts downwards.
This method of representing forces is known as Bow’s
notation.
To analyse the truss, we must first consider the joint
ABD ; this is because this joint has only two unknown
forces, namely the internal forces in the two members
that meet at the joint ABD. Neither joints BCD and
CAD can be considered first, because each of these
joints has more than two unknown forces.
Now the 3 kN load is between the spaces A and B ,
so that it can be represented by the lower case letters
ab , point from a to b and of magnitude 3 kN, as shown
in Figure 7.9.
a
3 kN
b
Figure 7.9
Similarly, the force in the truss between the spaces B
and D, namely the vector bd , lies at 60° to the hori¬
zontal and the force in the truss between the spaces D
and A, namely the vector da , lies at 30° to the horizon¬
tal. Thus, in Figure 7.9, if the vectors bd and ad are
drawn, they will cross at the point d , where the point d
will obviously lie to the left of the vector ab , as shown.
Hence, if the vector ab is drawn to scale, the magni¬
tudes of the vectors bd and da can be measured from
the scaled drawing. The direction of the force in the
member between the spaces B and D at the joint ABD
point upwards because the vector from b to d points
upwards. Similarly, the direction of the force in the
member between the spaces D and A at the joint ABD
is also upwards because the vector from d to a points
upwards. These directions at the joint ABD are shown
in Figure 7.10. Now as the framework is in equilib¬
rium, the internal forces in the members BD and DA ,
at the joints (2) and (1) respectively, will be equal and
opposite to the internal forces at the joint ABD ; these
are shown in Figure 7.10.
Figure 7.10
Comparing the directions of the arrows in
Figure 7.10 with those of Figure 7.3, it can be seen that
Forces in structures 101
the members BD and DA are in compression and are
defined as struts. It should also be noted from Figure
7.10, that when a member of the framework, say, BD ,
is so defined, we are referring to the top joint, because
we must always work around a joint in a clock¬
wise manner; thus the arrow at the top of BD points
upwards, because in Figure 7.9, the vector bd points
upwards from b to d. Similarly, if the same member is
referred to as DB , then we are referring to the bottom
of this member at the joint (2), because we must always
work clockwise around a joint. Hence, at joint (2), the
arrow points downwards, because the vector db points
downwards from d to b in Figure 7.9.
To determine the unknown forces in the horizontal
member between joints (1) and (2), either of these
joints can be considered, as both joints now only have
two unknown forces. Let us consider joint (1), i.e. joint
ADC. Now the vector ad can be measured from Figure
7.9 and drawn to scale in Figure 7.11.
Figure 7.11
Now the unknown force between the spaces D and C,
namely the vector dc , is horizontal and the unknown
force between the spaces C and A, namely the vector
ca , is vertical, hence, by drawing to scale and direction,
the point c can be found. This is because the point c in
Figure 7.11 lies below the point a and to the right of d.
In Figure 7.11, the vector ca represents the magnitude
and direction of the unknown reaction R j and the vector
dc represents the magnitude and direction of the force in
the horizontal member at joint (1); these forces are shown
in Figure 7.12, where R x = 0.82 kN and dc = 1.25 kN.
Figure 7.12
Comparing the directions of the internal forces in the
bottom of the horizontal member with Figure 7.3, it can
be seen that this member is in tension and, therefore, it
is a tie.
The reaction R 2 can be determined by consider¬
ing joint (2), i.e. joint BCD , as shown in Figure 7.13,
where the vector be represents the unknown reaction
R 1 which is measured as 2.18 kN.
Figure 7.13
The complete vector diagram for the whole frame¬
work is shown in Figure 7.14, where it can be seen that
R1 + R2 = 3 kN, as required by the laws of equilibrium.
It can also be seen that Figure 7.14 is a combination
of the vector diagrams of Figures 7.9, 7.11 and 7.13.
Experience will enable this problem to be solved more
quickly by producing the vector diagram of Figure 7.14
directly.
Figure 7.14
The table below contains a summary of all the mea¬
sured forces.
Member
Force (kN)
bd
-2.6
da
-1.55
cd
1.25
0.82
r 2
2.18
Determine the internal forces in
the members of the truss of Figure 7.15 due to the
externally applied horizontal force of 4 kN at the
joint ABE.
Part Two
Part Two
102 Mechanical Engineering Principles
In this case, the spaces between the unknown forces
are A, B, C, D and E. It should be noted that the reac¬
tion at joint (1) is vertical because the joint is on roll¬
ers, and that there are two reactions at joint (2) because
it is firmly anchored to the ground and there is also a
horizontal force of 4 kN which must be balanced by
the unknown horizontal reaction H 2 . If this unknown
horizontal reaction did not exist, the structure would
‘float’ into space due to the 4 kN load.
Consider joint ABE , as there are only two unknown
forces here, namely the forces in the members BE and
EA. Working clockwise around this joint, the vector
diagram for this joint is shown in Figure 7.16.
Figure 7.16
Joint (2) cannot be considered next, as it has three un¬
known forces, namely // 2 , R 2 and the unknown mem¬
ber force DE. Hence, joint (1) must be considered next;
it has two unknown forces, namely R l and the force in
member ED. As the member AE and its direction can
be obtained from Figure 7.16, it can be drawn to scale
in Figure 7.17. By measurement, de = 3 kN.
a
Figure 7.17
As R x is vertical, then the vector da is vertical, hence,
the position d can be found in the vector diagram of
Figure 7.17, where Ry= da (pointing downwards).
To determine R 1 and // 2 , joint (2) can now be con¬
sidered, as shown by the vector diagram for the joint in
Figure 7.18.
d 3 kN e
c
Figure 7.18
The complete diagram for the whole framework is
shown in Figure 7.19, where it can be seen that this
diagram is the sum of the vector diagrams of Figures
7.16 to 7.18.
Figure 7.19
The table below contains a summary of all the mea¬
sured forces
Member
Force (kN)
be
-2.1
ae
3.5
de
-3.0
-1.8
*2
1.8
h 2
4.0
Couple and moment
Prior to solving Problem 7, it will be necessary for
the reader to understand the nature of a couple; this is
described in Chapter 11, page 151.
The magnitude of a couple is called its moment; this is
described in Chapter 6, page 86.
Determine the internal forces in the
pin-jointed truss of Figure 7.20.
3 kN
Figure 7.20
Forces in structures 103
In this case, there are more than two unknowns at every
joint; hence it will first be necessary to calculate the
unknown reactions R { and R 2
To determine R v take moments about joint (2):
Clockwise moments about joint (2) = counter-clockwise
(or anticlockwise) moments about joint (2)
i.e. 7* 1 x8m = 4kNx6m + 3kNx4m + 5kNx2m
= 24+ 12+ 10 = 46 kNm
Therefore, R x = 46kNm = 5.75 kN
1 8m
Resolving forces vertically:
Upward forces = downward forces
i.e. 7? 1 +7? 2 = 4 + 3 + 5 = 12 kN
However, R { = 5.75 kN, from above,
hence, 5.75 kN + R 2 = 12 kN
from which, R 2 = 12 - 5.75 = 6.25 kN
Placing these reactions on Figure 7.21, together with
the spaces between the lines of action of the forces, we
can now begin to analyse the structure.
Figure 7.21
Starting at either joint AFE or joint DEJ , where there
are two or less unknowns, the drawing to scale of the
vector diagram can commence. It must be remembered
to work around each joint in turn, in a clockwise man¬
ner, and only to tackle a joint when it has two or less
unknowns. The complete vector diagram for the entire
structure is shown in Figure 7.22.
The table below contains a summary of all the mea
sured forces.
Member
Force (kN)
of
-11.5
fe
10.0
jd
-12.5
ej
10.9
bg
-7.5
gf
-4.0
ch
-7.6
hg
4.6
jh
-5.0
5.75
*2
6.25
Now try the following Practice Exercise
Practice Exercise 46 Further problems on
a graphical method
Determine the internal forces in the following
pin-jointed trusses using a graphical method:
Figure 7.23
7^ = 3.0 kN, 7* 2 = 1.0 kN,
1-2, 1.7 kN, 1 -3,-3.5kN,
,2 - 3, - 2.0 kN
Figure 7.22
Part Two
Part Two
104 Mechanical Engineering Principles
2 .
'R { = -2.6kN,7? 2 = 2.6kN,
H 2 = 6. OkN, 1-2,- 1.5 kN,
.1 -3,3.0kN,2-3,-5.2kN
6 kN
Figure 7.25
R^ = 5.0 kN, 7?-, = 1.0 kN,
H l = 4.0 kN, 1-2, 1.0 kN,
,1 -3,-7.1 kN, 2 - 3, - 1.4 kN
4 kN
Figure 7.26
= 5.0 kN, R 2 = 7.0 kN, 1 - 3, -10.0 kN, '
1 - 6, 8.7 kN, 3-4, -8.0 kN, 3 - 6,
-2.0 kN, 4 - 6, 4.0 kN, 4 - 5, -8.0 kN, 5-6,
-6.0 kN, 5-2, -14.0 kN, 6-2, 12.1 kN
7.4 Method of joints
(a mathematical method)
In this method, all unknown internal member forces are
initially assumed to be in tension. Next, an imaginary
cut is made around a joint that has two or less unknown
forces, so that a free body diagram is obtained for this
joint. Next, by resolving forces in respective vertical
and horizontal directions at this joint, the unknown
forces can be calculated. To continue the analysis,
another joint is selected with two or less unknowns and
the process repeated, remembering that this may only
be possible because some of the unknown member
forces have been previously calculated. By selecting, in
turn, other joints where there are two or less unknown
forces, the entire framework can be analysed.
It must be remembered that if the calculated force in a
member is negative, then that member is in compres¬
sion. Vice-versa is true for a member in tension.
To demonstrate the method, some pin-jointed trusses
will now be analysed in Problems 8 to 10.
Solve Problem 5, Figure 7.8 on page
100, by the method of joints.
Firstly, assume all unknowns are in tension, as shown
in Figure 7.27.
Next, make imaginary cuts around the joints, as shown
by the circles in Figure 7.27. This action will give
us three free body diagrams. The first we consider
is around joint (1), because this joint has only two
unknown forces; see Figure 7.28.
Figure 7.28
Forces in structures 105
Resolving forces horizontally at joint (1):
Forces to the left = forces to the right
i.e.
i.e.
0.5 F 2
F x cos 30° = F 2 cos 60°
0.866 F x = 0.5 F 2
from which,
i.e.
Fl 0.866
F t = 0.577 F 2
(7.1)
Resolving forces vertically at joint (1):
Upward forces = downward forces
i.e. 0 = 3 kN + F x sin 30° + F 2 sin 60°
i.e. 0 = 3 + 0.5 Fj + 0.866 F 2 (7.2)
Substituting equation (7.1) into equation (7.2) gives:
i.e.
0 = 3 + 0.5 x 0.577 F 2 + 0.866 F 2
— 3 = 1.1545 is
from which,
Fl 1.1545
i.e.
F 2 = -2.6 kN (compressive) (7.3)
Substituting equation (7.3) into equation (7.1) gives:
F x = 0.577 x (- 2.6)
i.e. F x - -1.5 kN (compressive)
Considering joint (2), as it now has two or less un¬
known forces; see Figure 7.29.
1.5 kN
Figure 7.29
Resolving horizontally:
Forces to the left = forces to the right
i.e. 0 = 7^ cos 30° + F 3
Member
Force (kN)
Fx
-1.5
f 2
-2.6
1.3
Solve Problem 6, Figure 7.15 on
page 102, by the method of joints.
Firstly, we will assume that all unknown internal forces
are in tension, as shown by Figure 7.30.
Figure 7.30
Next, we will isolate each joint by making imagi¬
nary cuts around each joint, as shown by the circles
in Figure 7.30; this will result in three free body dia¬
grams. The first free body diagram will be for joint
(1), as this joint has two or less unknown forces; see
Figure 7.31.
4 kN
Figure 7.31
Resolving forces horizontally:
However, F x =
-1.5 kN,
F x cos 30°
hence,
0 = -1.5 x 0.866 + F,
j
i.e.
0.866 F x
from which,
F 3 = 1.30 kN (tensile)
These results
are similar to those obtained by the
from which, F j
graphical method used in Problem 5; see Figure 7.12
on page 101, and the table at the top of the next column.
or
Fx
= 4 kN + F 9 cos 60°
= 4 + 0.5 F 2
4 + 0.5 F 2
~ 0.866
(7.4)
Part Two
Part Two
106 Mechanical Engineering Principles
Resolving forces vertically:
i.e.
or
or
0 = Fj sin 30° + F 2 sin 60°
0 = 0.5 Fj + 0.866 F 2
~Fi =
0.8 66F 2
0.5
Fj= -1.732 F 2
(7.5)
Equating equations (7.4) and (7.5) gives:
4.619 kN + 0.577 F 2 = -1.732 F 2
i.e.
4.619 =-1.732 F 2 - 0.577 F 2
= -2.309 F,
Hence
4.619
2 2.309
i.e. F 2 = -2 kN (compressive) (7.6)
Substituting equation (7.6) into equation (7.5) gives:
Fj = -1.732 x (-2)
i.e. Fj = 3.465 kN (7.7)
Consider next joint (2), as this joint now has two or
less unknown forces; see Figure 7.32.
Figure 7.32
Resolving forces vertically:
R x +Fj sin 30° = 0
or R \=-F l sin 30° (7.8)
Substituting equation (7.7) into equation (7.8) gives:
R { = -3.465x0.5
i.e. Fj = -1.733 kN (acting downwards)
Resolving forces horizontally:
0 = Fj cos30° + F 3
or F 3 = -F { cos 30° (7.9)
Substituting equation (7.7) into equation (7.9) gives:
F 3 = -3.465 x 0.866
i.e. F 3 = -3 kN (compressive) (7.10)
Consider next joint (3); see Figure 7.33.
Figure 7.33
Resolving forces vertically:
F 2 sin 60° + R 2 = 0
i.e. R 2 = —F 2 sin 60° (7.11)
Substituting equation (7.6) into equation (7.11) gives:
R 2 = -(-2) x 0.866
i.e. R 2 = 1.732 kN (acting upwards)
Resolving forces horizontally:
F 3 + F ? cos 60° + H 2 = 0
i.e. H 2 = -F 3 -F 2 x 0.5 (7.12)
Substituting equations (7.6) and (7.10) into equation
(7.12) gives:
H 2 = -(-3) - (-2) x 0.5
i.e. H 2 — 4 kN
These calculated forces are of similar value to those
obtained by the graphical solution for Problem 6, as
shown in the table below.
Member
Force (kN)
3.47
F 2
-2.0
F,
-3.0
-1.73
R 2
1.73
1-U
4.0
Solve Problem 7, Figure 7.20 on
page 102, by the method of joints.
Firstly, assume all unknown member forces are in
tension, as shown in Figure 7.34.
Forces in structures 107
Joint®
Figure 7.34
Next, we will isolate the forces acting at each joint by
making imaginary cuts around each of the five joints as
shown in Figure 7.34.
As there are no joints with two or less unknown
forces, it will be necessary to calculate the unknown
reactions R x and R 2 prior to using the method of joints.
Using the same method as that described for the
solution of Problem 7, we have
R x = 5.75 kN and R 2 = 6.25 kN
Now either joint (1) or joint (2) can be considered, as
each of these joints has two or less unknown forces.
Consider joint (1); see Figure 7.35.
Figure 7.35
Resolving forces vertically:
5.75 + F x sin 30° = 0
i.e.
or
i.e.
i.e.
F, sin 30° = -5.75
l
0.5 F x - -5.75
Z7 - 5 - 75
" 0.5
F x =-11.5 kN (compressive) (7.13)
Resolving forces horizontally:
i.e.
0 = F 2 + F { cos 30°
F 2 = -F x cos 30°
(7.14)
Substituting equation (7.13) into equation (7.14)
gives:
F 2 = -F x cos 30° = -(-11.5) x 0.866
i.e. F 2 = 9.96 kN (tensile)
Consider joint (2); see Figure 7.36.
Fa
Figure 7.36
Resolving forces vertically:
R ? + F 4 sin 30° = 0
i.e.
or
R 2 + 0.5 F 4 = 0
F= -*2
4 0.5
6.25
(7.15)
Since R 0 = 6.25, F d = - ^ _
z H 0.5
F 4 = -12.5 kN (compressive)
i.e.
(7.16)
Resolving forces horizontally:
F 3 + / 7 4 cos30° = 0
i.e. f 3 = -F 4 cos 30° (7.17)
Substituting equation (7.16) into equation (7.17) gives:
F 3 = -(-12.5) x 0.866
i.e. F 3 = 10.83 kN (tensile)
Consider joint (3); see Figure 7.37.
Figure 7.37
Resolving forces vertically:
F 6 sin 30° = F x sin 30° + F 5 sin 30° + 4
i.e.
^6 = ^ 1 +^ +
sin 30°
(7.18)
Part Two
Part Two
108 Mechanical Engineering Principles
Substituting equation (7.13) into equation (7.18)
gives:
F 6 = - 11.5+F 5 + 8 (7.19)
Resolving forces horizontally:
F x cos 30° = F 5 cos 30° + F 6 cos 30°
i.e. Fi=F 5 +F 6 (7.20)
Substituting equation (7.13) into equation (7.20) gives:
-11.5=F 5 + j F 6
or F 6 = -11.5-F 5 (7.21)
Equating equations (7.19) and (7.21) gives:
- 11.5 + F 5 + 8 = -11.5 -F 5
or F 5 +F 5 = -11.5 + 11.5-8
i.e. 2 F 5 = -8
from which, F 5 = -4kN (compressive) (7.22)
Substituting equation (7.22) into equation (7.21) gives:
F 6 = -11.5-(-4)
i.e. F 6 = -7.5 kN (compressive) (7.23)
Consider joint (4); see Figure 7.38.
3 kN
Resolving forces horizontally:
F 6 cos 30° = F g cos 30°
i.e. F 6 = F 8
but from equation (7.23), F 6 = -7.5 kN
Hence, F g = -7.5 kN (compressive) (7.24)
Resolving forces vertically:
0 = 3 + F 6 sin 30° + F 1 + F g sin 30°
i.e. 0 = 3+0.5F 6 +F ? + 0.5F 8 (7.25)
Substituting equations (7.23) and (7.24) into equation
(7.25) gives:
0 = 3 + 0.5 x - 7.5 + F 7 + 0.5 x - 7.5
or F 1 = -3 + 0.5 x 7.5 + 0.5 x 7.5
= -3 + 7.5
from which, F 1 = 4.5 kN (tensile) (7.26)
Consider joint (5); see Figure 7.39.
F 8 = -7.5 kN
12.5 kN
Figure 7.39
Resolving forces horizontally:
F g cos 30° + F 9 cos 30° = F 4 cos 30°
i.e. Fg + F 9 = F 4 (7.27)
Substituting equations (7.24) and (7.16) into equation
(7.27) gives:
- 7.5 + F 9 = -12.5
i.e. F 9 =-12.5 + 7.5
i.e. F 9 =-5 kN (compressive)
The results compare favourably with those obtained by
the graphical method used in Problem 7; see the table
below.
Member
Force (kN)
F \
-11.5
F 2
9.96
F 2
10.83
F 4
-12.5
F 5
-4.0
F 6
-7.5
F 1
4.5
F *
-7.5
F 9
-5.0
Forces in structures 109
Now try the following Practice Exercise
Practice Exercise 47 Further problems on
the method of joints
Using the method of joints, determine the
unknown forces for the following pin-jointed
trusses:
1. Figure 7.23 (page 103)
R { = 3.0 kN, R 2 = 1.0 kN, 1 - 2,
1.73 kN, 1 - 3, -3.46 kN, 2-3, -2.0 kN
2. Figure 7.24 (page 104)
Rj = -2.6 kN, R 2 = 2.6 kN,
H 2 = 6.0 kN, 1 -2,- 1.5 kN,
.1-3,3.0 kN, 2-3,-5.2 kN
3. Figure 7.25 (page 104)
R { =5.0kN,R 2 = 1.0 kN,
H { = 4.0 kN, 1-2, 1.0 kN, 1-3,
-7.07 kN, 2-3,- 1.41 kN
4. Figure 7.26 (page 104)
R { = 5.0 kN, R 2 - 7.0 kN, 1-3,
- 10.0 kN, 1—6, 8.7 kN, 3-4,
-8.0 kN, 3-6, -2.0 kN, 4-6,
4.0 kN, 4-5,-8.0 kN, 5 - 6,
- 6 kN, 5 - 2, -14 kN, 2-6, 12.1 kN
Firstly, all members will be assumed to be in tension
and an imaginary cut will be made through the frame¬
work, as shown by Figure 7.40.
|3kN
Figure 7.40
Taking moments about B; see Figure 7.41.
7.5 The method of sections
(a mathematical method)
In this method, an imaginary cut is made through
the framework and the equilibrium of this part of the
structure is considered through a free body diagram.
No more than three unknown forces can be determined
through any cut section, as only three equilibrium con¬
siderations can be made, namely
(a) resolve forces horizontally
(b) resolve forces vertically
(c) take moments about a convenient point.
Worked problem 11 demonstrates the method of sec¬
tions.
Determine the unknown member
forces F 2 , F 5 and F 6 of the truss of Figure 7.34,
Problem 10, page 107, by the method of sections.
Figure 7.41
Clockwise moments = anti-clockwise moments
Hence, 5.75 kN x 2 m = F 2 x 1.155 m
(where 2 tan 30° = 1.155 m from Figure 7.41)
i.e.
5.75x2
F 2 = 1 155 = 9.96 kN (tensile)
(7.28)
Resolving forces vertically:
5.75 kN + F 6 sin 30° = F 5 sin 30° + 4 kN
i.e.
i.e.
77 _ 77 , 5.75
F 5 - F 6 + 0.5
F s = F 6 + 3.5
0.5
(7.29)
Resolving forces horizontally:
0 = F 2 + F 5 cos 30° + F e cos 30°
from which, F 5 cos 30° = -F 2 ~ F 6 cos 30°
and
^5 = - —
- F.
6
cos30°
(7.30)
Part Two
Part Two
110 Mechanical Engineering Principles
Substituting equation (7.28) into equation (7.30) gives:
9.96
F 5 = -
0.866 F(<
or F 5 = -11.5-F 6 (7.31)
Equating equation (7.29) to equation (7.31) gives:
F 6 + 3.5=-11.5- j F 6
from which, 2 F 6 = —11.5 - 3.5 = —15
and F 6 = ~ -y = -7.5 kN (compressive) (7.32)
Substituting equation (7.32) into equation (7.31) gives:
F 5 =-11.5-(-7.5)
= -4 kN (compressive) (7.33)
i.e. F 2 = 9.96 kN, F 5 = -4 kN and F 6 = -7.5 kN
The above answers can be seen to be the same as those
obtained in Problem 10.
Now try the following Practice Exercises
Practice Exercise 48 Further problems
on the method of
sections
Determine the internal member forces of the fol¬
lowing trusses, by the method of sections:
1. Figure 7.23 (page 103)
= 3.0 kN, R 2 = 1.0 kN,
1-2, 1.73 kN, 1-3,
-3.46 kN, 2 - 3, - 2.0 kN
2. Figure 7.24 (page 104)
R { = -2.6 kN, R 2 = 2.6 kN,
H 2 = 6 kN, 1 -2,- 1.5 kN,
.1 -3,3.0kN,2-3,-5.2kN.
3. Figure 7.25 (page 104)
R x = 5.0 kN, R~,= 1.0 kN,
H { = 4.0 kN, 1-2, 1.0 kN,
.1 -3,-7.07 kN, 2-3,-1.41 kN
4. Figure 7.26 (page 104)
R { = 5.0 kN, R 2 = 7.0 kN, 1-3,
-10.0 kN, 1-6, 8.7 kN, 3-4,
-8.0 kN, 3-6, -2.0 kN, 4-6,
4.0 kN, 4-5, -8.0 kN, 5-6,
-6.0 kN, 5 - 2, -14.0 kN, 6 - 2, 12.1 kN
Practice Exercise 49 Short-answer
questions on
forces in structures
1. Where must the loads be applied on a pin-
jointed truss?
2. If there are three unknown forces in a truss,
how many simultaneous equations are re¬
quired to determine these unknown forces?
3. When is a truss said to be statically indeter¬
minate?
4. For a plane pin-jointed truss, what are the
maximum number of unknowns that can ex¬
ist at a joint to analyse that joint without ana¬
lysing another joint before it?
Practice Exercise 50 Multiple-choice
questions on forces in
frameworks
(Answers on page 335)
1. The truss of Figure 7.42 is a:
(a) mechanism
(b) statically determinate
(c) statically indeterminate
2. The value of F { in Figure 7.43 is:
(a) 1 kN (b) 0.5 kN
(c) 0.707 kN
1 kN
''
Figure 7.43
Forces in structures 111
3. The value of F 2 in Figure 7.43 is:
(a) 0.707 kN (b) 0.5 kN
(c) 0
4. If the Young’s modulus is doubled in the
members of a pin-jointed truss, and the ex¬
ternal loads remain the same, the internal
forces in the truss will:
(a) double
(b) halve
(c) stay the same
5. If the Young’s modulus is doubled in the
members of a pin-jointed truss, and the loads
remain the same, the deflection of the truss
will:
(a) double (b) halve
(c) stay the same
6. If the external loads in a certain truss are
doubled, the internal forces will:
(a) double (b) halve
(c) stay the same
References
[1] CASE, J, LORD CHILVER and ROSS, C.T.F. Strength
of Materials and Structures Butterworth/Heinemann,
1999.
[2] ROSS, C.T.F. Mechanics of Solids , Elsevier/Oxford,
U.K. 1999.
For fully worked solutions to each of the problems in Practice Exercises 46 to 50 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Two
Chapter 8
Why it is important to understand: Bending moment and shear force diagrams
The members of the structures in the previous chapter withstood the externally applied loads in either
tension or compression; this was because they were not subjected to bending. In Practice many structures
are subjected to bending action; such structures include beams and rigid-jointed frameworks (a rigid-jointed
framework is one which has its joints welded or riveted or bolted together; such structures are beyond the
scope of this text - see reference [1], on page 111). In this chapter, bending moments and shearing forces are
defined and calculated, before the plotting of bending moment and shearing force diagrams are explained.
Bending moments and shearing forces are of much importance in the design of buildings, bridges, ships, etc.
At the end of this chapter you should be able to:
• define a rigid-jointed framework
• define bending moment
• define sagging and hogging
• define shearing force
• calculate bending moments
• calculate shearing forces
• plot bending moment diagrams
• plot shearing force diagrams
• define the point of contraflexure
8.1 Bending moment (M)
The units of bending moment are N mm, N m, kN m,
etc. When a beam is subjected to the couples shown in
Figure 8.1, the beam will suffer flexure due to the bend¬
ing moment of magnitude M.
M
M
(b) Hogging moment (-)
Figure 8.1 Bending moments
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
Bending moment and shear force diagrams 113
If the beam is in equilibrium and it is subjected to
a clockwise couple of magnitude M on the left of the
section, then from equilibrium considerations, the
couple on the right of the section will be of exactly
equal magnitude and of opposite direction to the cou¬
ple on the left of the section. Thus, when calculating
the bending moment at a particular point on a beam
in equilibrium, we need only calculate the magnitude
of the resultant of all the couples on one side of the
beam under consideration. This is because as the beam
is in equilibrium, the magnitude of the resultant of all
the couples on the other side of the beam is exactly
equal and opposite. The beam in Figure 8.1(a) is said
to be sagging and the beam in Figure 8.1(b) is said to
be hogging.
The sign convention adopted in this text is:
(a) sagging moments are said to be positive
(b) hogging moments are said to be negative
8.2 Shearing force (F)
Whereas a beam can fail due to its bending moments be¬
ing excessive, it can also fail due to other forces being too
large, namely the shearing forces; these are shown in Fig¬
ure 8.2. The units of shearing force are N, kN, MN, etc.
(a) Positive shearing force (b) Negative shearing force
Figure 8.2 Shearing forces
It can be seen from Figures 8.2(a) and (b) that the
shearing forces F act in a manner similar to that ex¬
erted by a pair of garden shears when they are used
to cut a branch of a shrub or a plant through shearing
action. This mode of failure is different to that caused
by bending action.
In the case of the garden shears, it is necessary for the
blades to be close together and sharp, so that they do
not bend the branch at this point. If the garden shears
are old and worn the branch can bend and may lie be¬
tween the blades. Additionally, if the garden shears are
not sharp, it may be more difficult to cut the branch
because the shearing stress exerted by the blades will
be smaller as the contact area between the blades and
the branch will be larger.
The shearing action is illustrated by the sketch of
Figure 8.3.
Figure 8.3 Shearing action
Once again, if the beam is in equilibrium, then the
shearing forces either side of the point being consid¬
ered will be exactly equal and opposite, as shown in
Figures 8.2(a) and (b). The sign convention for shear¬
ing force is that it is said to be positive if the right
hand is going down; see Figure 8.2(a).
Thus, when calculating the shearing force at a par¬
ticular point on a horizontal beam, we need to calculate
the resultant of all the vertical forces on one side of
the beam, as the resultant of all the vertical forces on
the other side of the beam will be exactly equal and op¬
posite. The calculation of bending moments and shear¬
ing forces and the plotting of their respective diagrams
are demonstrated in the following worked problems.
8.3 Worked problems on
bending moment and
shearing force diagrams
Calculate and sketch the bending
moment and shearing force diagrams for the
horizontal beam shown in Figure 8.4, which is
simply supported at its ends.
6 kN
A
B
3 m
Figure 8.4
Firstly, it will be necessary to calculate the magnitude
of reactions R 4 and R B
Taking moments about B gives:
Clockwise moments about B =
anticlockwise moments about B
Part Two
Part Two
114 Mechanical Engineering Principles
i.e. R 4 x 5 m = 6 kN x2m= 12 kN m
12
from which, R { - — =2.4 kN
Resolving forces vertically gives:
Upward forces = downward forces
i.e. R a + R b = 6 kN
i.e. 2A + R b = 6
from which, R B - 6 - 2.4 = 3.6 kN
As there is a discontinuity at point C in Figure 8.4, due
to the concentrated load of 6 kN, it will be necessary to
consider the length of the beam AC separately from the
length of the beam CB. The reason for this is that the
equations for bending moment and shearing force for
span AC are different to the equation for the span CB;
this is caused by the concentrated load of 6 kN.
For the present problem, to demonstrate the nature
of bending moment and shearing force, these values
will be calculated on both sides of the point of the beam
under consideration. It should be noted that, normally,
the bending moment and shearing force at any point on
the beam are calculated only due to the resultant cou¬
ples or forces, respectively, on one side of the beam.
Consider span AC:
Bending moment
Consider a section of the beam at a distance x from the
left end A, where the value of x lies between A and C,
as shown in Figure 8.5.
Figure 8.5
From Figure 8.5, it can be seen that the reaction R 4
causes a clockwise moment of magnitude R 4 x x = 2.4x
on the left of this section and as shown in the lower
diagram of Figure 8.5. It can also be seen from the up¬
per diagram of Figure 8.5, that the force on the right of
this section on the beam causes an anticlockwise mo¬
ment equal to R B x (5 - x) or 3.6(5 - x) and a clockwise
moment of 6 x (3 - x), resulting in an anticlockwise
moment of:
3.6(5 -x) - 6(3 -x) = 3.6 x 5 - 3.6x - 6 x 3 + 6x
= 18 — 3.6x — 18 + 6x
= 2.4x
Thus, the left side of the beam at this section is sub¬
jected to a clockwise moment of magnitude 2.4x and
the right side of this section is subjected to an anti¬
clockwise moment of 2.4x, as shown by the lower
diagram of Figure 8.5. As the two moments are of
equal magnitude but of opposite direction, they cause
the beam to be subjected to a bending moment M -
2 Ax. As this bending moment causes the beam to sag
between A and C, the bending moment is assumed to be
positive, or at any distance x between A and C:
Bending moment = M = +2.4x (8.1)
Shearing force
Here again, because there is a discontinuity at C, due
to the concentrated load of 6 kN at C, we must con¬
sider a section of the beam at a distance x from the left
end A, where x varies between A and C, as shown in
Figure 8.6.
F = 2.4 kN
Figure 8.6
From Figure 8.6, it can be seen that the resultant of
the vertical forces on the left of the section at x are
2.4 kN acting upwards. This force causes the left
of the section at x to slide upwards, as shown in the
lower diagram of Figure 8.6. Similarly, if the vertical
forces on the right of the section at x are considered,
it can be seen that the 6 kN acts downwards and that
R b = 3.6 kN acts upwards, giving a resultant of 2.4 kN
acting downwards. The effect of the two shearing forc¬
es acting on the left and the right of the section at x,
causes the shearing action shown in the lower diagram
of Figure 8.6. As this shearing action causes the right
side of the section to glide downwards, it is said to be a
positive shearing force.
Summarising, at any distance x between^ and C:
F = shearing force = +2.4 kN (8.2)
Bending moment and shear force diagrams 115
Consider span CB:
Bending moment
At any distance x between C and B , the resultant mo¬
ment caused by the forces on the left of x is given by:
M = R 4 xx — 6(x — 3) = 2Ax — 6(x — 3)
= 2 Ax — 6x + 18
i.e. M- 18 - 3.6x (clockwise) (8.3)
The effect of this resultant moment on the left of x is
shown in the lower diagram of Figure 8.7.
3 rrr
A
2 ,
6 kN
—
-2 m-
kN
B
\
3.6 kN
M = 2.4x
-6 (x - 3)
= 18 - 3.6x
M = 3.6 (5 - x)
= 18-3.6x
Figure 8.7
Now from Figure 8.7, it can be seen that on the right
side of x, there is an anticlockwise moment of:
M = R b x (5 -x) = 3.6(5 -x) = 18 - 3.6x
i.e. M= 18 - 3.6x (anticlockwise) (8.4)
The effect of the moment of equation (8.3) and that
of the moment of equation (8.4), is to cause the beam
to sag at this point as shown by the lower diagram of
Figure 8.7, i.e. M is positive between C and B , and
M=+18-3.6x (8.5)
Sh earing force
Consider a distance x between C and B , as shown in
Figure 8.8.
-3 m-
6 kN
C 1
-2 m
Figure 8.8
From Figure 8.8, it can be seen that at x, there are
two vertical forces to the left of this section, namely
the 6 kN load acting downwards and the 2.4 kN load
acting upwards, resulting in a net value of 3.6 kN act¬
ing downwards, as shown by the lower diagram of
Figure 8.8. Similarly, by considering the vertical forc¬
es acting on the beam to the right of x, it can be seen
that there is one vertical force, namely the 3.6 kN load
acting upwards, as shown by the lower diagram of
Figure 8.8. Thus, as the right hand of the section is
tending to slide upwards, the shearing force is said to
be negative, i.e. between C and B ,
F = - 3.6kN (8.6)
It should be noted that at C, there is a discontinuity
in the value of the shearing force, where over an infini¬
tesimal length the shearing force changes from +2.4 kN
to - 3.6 kN, from left to right.
Bending moment and shearing force diagrams
The bending moment and shearing force diagrams are
simply diagrams representing the variation of bending
moment and shearing force, respectively, along the length
of the beam. In the bending moment and shearing force
diagrams, the values of the bending moments and shear¬
ing forces are plotted vertically and the value of x is plot¬
ted horizontally, where x = 0 at the left end of the beam
and x = the whole length of the beam at its right end.
In the case of the beam of Figure 8.4, bending
moment distribution between^ and C is given by equa¬
tion (8.1), i.e. M = 2.4x, where the value of x varies
between A and C.
At A, x — 0, therefore M A = 2.4 x 0 = 0
and at C, x = 3 m, therefore M c - 2.4 x 3 = 7.2 kN m
Additionally, as the equation M = 2.4x is a straight line,
the bending moment distribution between A and C will
be as shown by the left side of Figure 8.9(a).
Figure 8.9 Bending moment and shearing force diagrams
Part Two
Part Two
116 Mechanical Engineering Principles
Similarly, the expression for the variation of bending
moment between C and B is given by equation (8.3),
i.e. M = 18 - 3.6x, where the value of x varies between
C and B. The equation can be seen to be a straight line
between C and B.
At C, x = 3 m,
therefore M c = 18-3.6x3= 18 - 10.8 = 7.2 kN m
At B, x = 5 m,
therefore M B — 18-3.6x5 = 18 - 18 = 0
Therefore, plotting of the equation M - 18 - 3.6x be¬
tween C and B results in the straight line on the right of
Figure 8.9(a), i.e. the bending moment diagram for this
beam has now been drawn.
In the case of the beam of Figure 8.4, the shearing
force distribution along its length from A to C is given
by equation (8.2), i.e. F = 2.4 kN, i.e. F is constant
between A and C. Thus the shearing force diagram
between A and C is given by the horizontal line shown
on the left of C in Figure 8.9(b).
Similarly, the shearing force distribution to the
right of C is given by equation (8.6), i.e. F = -3.6 kN,
i.e. F is a constant between C and B , as shown by the
horizontal line to the right of C in Figure 8.9(b). At
the point C, the shearing force is indeterminate and
changes from +2.4 kN to -3.6 kN over an infinitesimal
length.
Determine expressions for the distri¬
butions of bending moment and shearing force for
the horizontal beam of Figure 8.10. Hence, sketch
the bending and shearing force diagrams.
5 kN
' >
6 kN
' V
A >
D >
IB
^^
2m
^
2m
^^
3 m
1 m
Ra Rb
Figure 8.10
Firstly, it will be necessary to calculate the unknown
reactions R 4 and R B
Taking moments about B gives:
R a x 5 m + 10 kN x 1 m = 5 kN x 7 m + 6 kN x 3 m
i.e. 5 7^+10 = 35+18
5 R a = 35 + 18-10 = 43
43
R a = — = 8.6 kN
Resolving forces vertically gives:
R a + R b = 5kN + 6kN+ 10 kN
i.e. 8.6+ 7?^ = 21
from which, R B = 21 - 8.6 = 12.4 kN
For the range C to A , see Figure 8.11.
/
/
/
/
/
5 kN
C
Figure 8.11
To calculate the bending moment distribution (M),
only the resultant of the moments to the left of the
section at x will be considered, as the resultant of the
moments on the right of the section of x will be exactly
equal and opposite.
Bending moment (BM)
From Figure 8.11, at any distance x,
M = -5 x x (hogging) = -5x (8.7)
Equation (8.7) is a straight line between C and A.
At C, x = 0, therefore M c = -5x0 = 0 kN m
At A, x = 2 m, therefore M { = -5x2 = -10 kN m
Shearing force (SF)
To calculate the shearing force distribution (F) at any
distance x, only the resultant of the vertical forces to
the left of x will be considered, as the resultant of the
vertical forces to the right of x will be exactly equal
and opposite.
From Figure 8.11, at any distance x,
F=- 5kN (8.8)
It is negative, because as the left of the section tends to
slide downwards, the right of the section tends to slide
upwards. (Remember, right hand down is positive.)
For the range A to D, see Figure 8.12.
from which,
Figure 8.12
Bending moment and shear force diagrams 117
Bending moment (BM)
At any distance x between A and D
M = -5 x x + R a x (x - 2)
= —5x + 8.6(x - 2)
= —5x + 8.6x - 17.2
i.e. 3.6*-17.2
(a straight line between A and D) (8.9)
At y 4, x = 2 m, M a = 3.6x2- 17.2
= 7.2 - 17.2 = -10 kN m
At Z), x = 4 m, M D = 3.6x4- 17.2
= 14.4 - 17.2 = -2.8 kN m
Shearing force (SF)
At any distance x between A and D, (8.10)
F = -5 kN + 8.6 kN = 3.6 kN (constant)
For the range D to B , see Figure 8.13.
15 kN
| 6 kN
A
R a = 8.6 kN
D
- ►
^ Z ill ^
■< -
^ Z III ^
-y-
Bending moment (BM)
In this case it will be convenient to consider only the
resultant of the couples to the right of x (remember
that only one side need be considered, and in this case,
there is only one load to the right of x).
At x, M= -10 x (8 -x) = -80 + 10* (8.13)
Equation (8.13) can be seen to be a straight line
between B and E.
At B, x = 7 m,
therefore M B = -80 + 10 x 7 = -10 kN m
At E, x = 8 m,
therefore M E = -80 + 10x8 = 0 kN m
Shearing force (SF)
Atx, F= +10 kN (constant) (8.14)
Equation (8.14) is positive because the shearing force
is causing the right side to slide downwards.
The bending moment and shearing force diagrams are
plotted in Figure 8.15 with the aid of equations (8.7)
to (8.14) and the associated calculations at C, A, D, B
and E.
Figure 8.13
Bending moment (BM)
Atx, M = — 5 x x + 8.6 x (x - 2) - 6 x (x- 4)
= —5x + 8.6x —17.2 — 6x + 24
i.e. M = -2.4* + 6.8 (8.11)
(a straight line between D and B)
At D, x = 4 m, therefore M D = -2.4 x 4 + 6.8
= -9.6 + 6.8
i.e. M D = -2.8 kN m
At B, x = 7 m, therefore M n = -2.4 x 7 + 6.8
= -16.8 + 6.8
i.e. M n = -10 kN m
Shearing force (SF)
Atx, F = -5 + 8.6 - 6 = -2.4 kN (constant) (8.12)
For the range B to E , see Figure 8.14.
10 kN
E
M A
F
0
+10
a +3.6
+3.6
C 4
D
B
—
-2.4
-2.4 °
+10
->-x
(b) SF diagram (kN)
Figure 8.15
Determine expressions for the bend¬
ing moment and shearing force distributions for the
beam of Figure 8.16. Hence, sketch the bending
moment and shearing force diagrams.
15 kN m 30 kN m 10 kN
Figure 8.1
Figure 8.14
Part Two
Part Two
118 Mechanical Engineering Principles
Firstly, it will be necessary to calculate the reactions
R 4 and R b
Taking moments about B gives:
15 kN m + R a x 5 m + 10 kN x 1 m = 30 kN m
i.e. 5 R A = 30 — 10 - 15 = 5
from which, R A = ^ =1 kN
Resolving forces vertically gives:
R a +R b = 10 kN
i.e. 1 + R b = 10
from which, R A =10-1=9 kN
For the span Cto^4, see Figure 8.17
15 kN m
/
/
/
/
/
I* Y
/
/
A >
Figure 8.17
Bending moment (BM)
Atx, M- 15 kN m (constant) (8.15)
Shearing force (SF)
At x = 0, F = 0 kN (8.16)
For the span A to D, see Figure 8.18.
15 kN m
A
1/
/
1 kN
x - ->
-5 £- III r
r a =
-< -
Figure 8.18
Bending moment (BM)
At x, M- 15 kN m + R A x (x - 2)
= 15 + l(x-2)
= 15 +x - 2
i.e. M = 13 + x (a straight line) (8.17)
At y 4, x = 2 m,
therefore M A = 13 + 2 = 15kNm
At D, x = 4 m,
therefore Mn - 13 + 4 = 17 kN m
U (r)
Note that M n means that M is calculated to the left of D.
U ir)
Shearing force (SF)
At x, F = 1 kN (constant) (8.18)
For the spanD to B , see Figure 8.19.
15 kN m
30 kN m
A* D
1 kN
O KV-» W 1 ^ O W-
_W-
^ z m z iTi ^
i
i
y
^ A
/
Figure 8.19
Bending moment (BM)
At x, M- 15 kN m + 1 kN m x (x - 2) - 30 kN m
= 15 + x - 2 - 30
i.e M - x - 17 (a straight line) (8.19)
At D, x = 4 m,
therefore M „ = 4 — 17 = -13 kN m
D
(+)
Note that M n means that M is calculated just to the
right of D.
(+)
At B, x = 7 m,
therefore M B = 7 - 17 = -10 kN m
*(-)
Shearing force (SF)
At x, F = -1 kN (constant) (8.20)
For the span B to E , see Figure 8.20.
15 kN m 10 kN
/
/
1
7
/
/
/
/
\x
E
-
-<-
X
— 8 m-
-►
-►
Figure 8.20
In this case we will consider the right of the beam as
there is only one force to the right of the section at x.
M=-10x(8-x) = -80 + lOx (a straight line)
At x, F = 10 kN (positive as the right hand is going
down, and constant).
Plotting the above equations, for the various spans,
results in the bending moment and shearing force
diagrams of Figure 8.21.
Bending moment and shear force diagrams 119
+17
Figure 8.21
Calculate and plot the bending
moment and shearing force distributions for the
cantilever of Figure 8.22.
5 kN
r
^ V w
m w
A ^
^ o
^i i i
Figure 8.22
In the cantilever of Figure 8.22, the left hand end is free
and the right hand end is firmly fixed; the right hand
end is called the constrained end.
Bending moment (BM)
Atx in Figure 8.22, M = -5 kN xi
i.e. M = -5x (a straight line) (8.21)
Shearing force (SF)
Atx in Figure 8.22, F =-5 kN (a constant) (8.22)
For equations (8.21) and (8.22), it can be seen that the
bending moment and shearing force diagrams are as
shown in Figure 8.23.
Mi
0 -
0 —
M = -5x^ '''■
(a) BM diagram
-10 kN m
F A
0 -
-5 kN -
0 —>■ x
-5 kN
(b) SF diagram
Figure 8.23
Determine the bending moment and
shearing force diagram for the cantilever shown
in Figure 8.24, which is rigidly constrained at the
end B.
^5 kN
10 kN
/
/
/
/
-2 m->-
C
■<— 1 m -
/
/
—►/
Figure 8.24
For the span A to C, see Figure 8.25.
Figure 8.25
Bending moment (BM)
At x, M - -5 kN x x
i.e. M- -5x (a straight line)
(8.23)
Shearing force (SF)
At x, F = -5 kN (constant)
(8.24)
For the span C to B , see Figure 8.26.
5 kN
10 kN
Figure 8.26
Bending moment (BM)
At x, M=-5 kN x x - 10 kN x (x - 2)
= -5x - lOx + 20
i.e. M- 20 - 15x (a straight line) (8.25)
At C, x = 2 m, therefore M c = 20 - 15x2
= 20-30
i.e. A/ c = -10 kN m
At B, x = 3 m, therefore M B = 20 - 15x3
= 20-45
i.e. M b = -25 kN m
Part Two
Part Two
120 Mechanical Engineering Principles
Shearing force (SF)
At x in Figure 8.26, F = -5 kN - 10 kN
i.e. F= -15 kN (constant) (8.26)
From equations (8.23) to (8.26) and the associated
calculations, the bending moment and shearing force
diagrams can be plotted, as shown in Figure 8.27.
(b) SF diagram (kN)
Figure 8.27
A uniform section, 6 m long, horizontal
beam is simply supported at its ends. If this beam is
subjected to two vertically applied downward loads,
one of magnitude 5 kN at 2 m from the left support,
and the other of magnitude 10 kN, 4 m from the left
support, calculate and plot the bending moment and
shear force diagrams.
The loaded beam is shown in Figure 8.28.
5 kN 10 kN
Taking moments about B:
R a x6 = 5x4+10x2
i.e. 6 R a = 40 and R A = ^ = 6.67 kN
R a + R b = 5 + 10 = 15 kN
hence R B = 15 - 6.67 = 8.33 kN
Bending moments
AX A, BM=0; At C, BM = R A x 2 = 6.67 x 2
= 13.34 kN m
At D, BM = R a x 4 - 5 x 2 = 6.67 x 4 - 5 x 2
= 16.68 kN m
At B,BM= 0
Shearing force
At A, SF = + R a = 6.67 kN
At C, SF= 6.67 - 5 = 1.67 kN
At Z), SF = 6.67 - 5 - 10 = -8.33 kN
At B , SF = -8.33 kN
The bending moment and shearing force diagram is
shown in Figure 8.29.
Figure 8.28
Figure 8.29
Bending moment and shearforce diagrams 121
Now try the following Practice Exercise
4.
Practice Exercise 51 Further problems
on bending moment
and shearing force
diagrams
Determine expressions for the bending moment
and shearing force distributions for each of
the following simply supported beams; hence,
or otherwise, plot the bending moment and
shearing force diagrams.
1 .
Cl
3kN
A
1m
A
1 m
Figure 8.30
[see Figure 8.45(a) on page 125]
2 .
C
4kN
*
k
2 m
>
1 m
Figure 8.31
1 kN
t '
4 kN
'
>
‘4
. 2m
, 2m
, 2m ^
Figure 8.33
[see Figure 8.45(d) on page 125]
5.
>
4 kN
r
6 kN m
c
_
D
1 m
1 m
1 m
Figure 8.34
[see Figure 8.45(e) on page 126]
6 .
Figure 8.35
[see Figure 8.45(b) on page 125]
3.
7.
[see Figure 8.45(f) on page 126]
A
'
1 kN
' \
4 kN
0 D 1
1 m
1 m
1 m
B
A
2 kN
6 kN m
—
C
1.5 m
1.5 m
B
Figure 8.32
Figure 8.36
[see Figure 8.45(c) on page 125]
[see Figure 8.45(g) on page 126]
Part Two
Part Two
122 Mechanical Engineering Principles
8. A horizontal beam of negligible mass is of
length 7 m. The beam is simply supported
at its ends and carries three vertical loads,
pointing in a downward direction. The first
load is of magnitude 3 kN and acts 2 m from
the left end, the second load is of magnitude
2 kN and acts 4 m from the left end, and
the third load is of magnitude 4 kN and acts
6 m from the left end. Calculate the bending
moment and shearing force at the points of
discontinuity, working from the left support
to the right support.
Bending moments (kN m): 0, 7.14, 8.28,
5.42, 0; Shearing forces (kN): 3.57, 3.57/
0.57, 0.57/-1.43, -1.43/-5.43, -5.43 .
8.4 Uniformly distributed loads
Uniformly distributed loads (UDL) appear as snow
loads, self-weight of the beam, uniform pressure loads,
and so on. In all cases they are assumed to be spread
uniformly over the length of the beam in which they
apply. The units for a uniformly distributed load are
N/m, kN/m, MN/m, and so on.
Worked problems 7 and 8 involve uniformly distrib¬
uted loads.
At any distance x in Figure 8.38,
M =-10 kN xx (8.27)
i.e. M- - 5x 2 (a parabola) (8.28)
In equation (8.27), the weight of the uniformly distrib¬
uted beam up to the point x is (10 x x). As the centre of
x
gravity of the UDL is at a distance of — from the right
end of Figure 8.38, M = -10x x j
The equation is negative because the beam is hogging.
At x = 0, M = 0
At x = 5 m, M- -10 x 5 x — =-125 kN m
Shearing force (SF)
At any distance x in Figure 8.38, the weight of the UDL
is (10 x x) and this causes the left side to slide down, or
alternatively the right side to slide up.
Hence, F = -10x (a straight line) (8.29)
At x = 0, F - 0
At x = 5 m, F = -10 x 5 = -50 kN
Plotting of equations (8.28) and (8.29) results in the
distributions for the bending moment and shearing
force diagrams shown in Figure 8.39.
Determine expressions for the bend¬
ing moment and shearing force distributions for the
cantilever shown in Figure 8.37, which is subjected
to a uniformly distributed load, acting downwards,
and spread over the entire length of the cantilever.
iv = 10 kN/m
< -x-►
< -5 m- >■
Figure 8.37
Bending moment (BM)
iv = 10 kN/m
<-x-►
(b) SF diagram
Figure 8.38
Figure 8.39
Bending moment and shear force diagrams 123
Determine expressions for the bend¬
ing moment and shearing force diagrams for the
simply supported beam of Figure 8.40. The beam
is subjected to a uniformly distributed load ( UDL )
of 5 kN/m, which acts downwards, and it is spread
over the entire length of the beam.
w = 5 kN/m
Figure 8.40
Firstly, it will be necessary to calculate the reactions
R a and R b . As the beam is symmetrically loaded, it is
evident that:
Taking moments about B gives:
Clockwise moments about B =
anticlockwise moments about B
kN
(8.30)
i.e.
R a x 6 m = 5
x 6 m x 3 m
m
= 90 kN m
(8.31)
from which, R A =
90
= 15 kN =R r
6 B
On the right hand side of equation (8.31), the term
5 kN/m x 6 m is the weight of the UDL , and the length
of 3 m is the distance of the centre of gravity of the
UDL from B.
Bending moment (BM)
At any distance x in Figure 8.41,
w = 5kN/m
A
R a =15 kN
Figure 8.41
n - kN x
M = R, xx- 5 —xxx —
A m2
(8.32)
i.e. M- 15x - 2.5x 2 (a parabola) (8.33)
On the right hand side of equation (8.32), the term:
(R a x x) is the bending moment (sagging) caused by
the reaction, and the term
/
v
X
kN
-XXX —
m 2
\
/
, which is
kN
hogging, is caused by the UDL , where (5 — x x) is
m
x
the weight of the UDL up to the point x, and — is the
distance of the centre of gravity of the UDL from the
right side of Figure 8.41.
At x = 0, M = 0
At x = 3 m, M = 15 x 3 - 2.5 x 3 2 = 22.5 kN m
At x = 6 m, M - 15 x 6 - 2.5 x 6 2 = 0
Shearing force (SF)
At any distance x in Figure 8.41,
F = R -5 — xx
A m
i.e. F = 15 - 5x (a straight line)
(8.34)
(8.35)
On the right hand side of equation (8.34), the term
/ kN 1
is the weight of the UDL up to the point x;
x x
m
v *" /
this causes a negative configuration to the shearing
force as it is causing the left side to slide downwards.
At x = 0, 15 kN
At x = 3 m, F = 15-5x3 = 0
Atx = 6m ,F= 15 - 5 x 6 = -15 kN
Plotting of equations (8.33) and (8.35) results in
the bending moment and shearing force diagrams of
Figure 8.42.
0
-15 kN
Figure 8.42
Part Two
Part Two
124 Mechanical Engineering Principles
Now try the following Practice Exercises
Practice Exercise 52 Further problems
on bending moment
and shearing force
diagrams
Determine expressions for the bending moment
and shearing force distributions for each of the
following simply supported beams; hence, plot
the bending moment and shearing force diagrams.
1. Figure 8.43(a)
6 kN/m
Figure 8.43 Simply supported beams
[see Figure 8.46(a) on page 126]
2. Figure 8.43(b)
[see Figure 8.46(b) on page 126]
Determine expressions for the bending mo¬
ment and shearing force distributions for
each of the following cantilevers; hence,
or otherwise, plot the bending moment and
shearing force diagrams.
3. Figure 8.44(a)
6 kN/m
-<-5 m- >
(a)
5 kN/m
< -9 m-►
(b)
Figure 8.44 Cantilevers
4. Figure 8.44(b)
[see Figure 8.47(b) on page 126]
Practice Exercise 53 Short-answer
questions on bending
moment and shearing
force diagrams
1. Define a rigid-jointed framework.
2. Define bending moment.
3. Define sagging and hogging.
4. State two practical examples of uniformly
distributed loads.
5. Show that the value of the maximum shear¬
ing force for a beam simply supported at its
ends, with a centrally placed load of 3 kN, is
1.5 kN.
6. If the beam in question 5 were of span 4 m,
show that its maximum bending moment is
3 kN m.
7. Show that the values of maximum bending
moment and shearing force for a cantilever
of length 4 m, loaded at its free end with a
concentrated load of 3 kN, are 12 kN m and
3 kN.
Practice Exercise 54 Multiple-choice
questions on bending
moment and shearing
force diagrams
(Answers on page 335)
1. A beam simply supported at its end, carries a
centrally placed load of 4 kN. Its maximum
shearing force is:
(a) 4 kN (b) 2 kN
(c) 8 kN (d) 0
2. Instead of the centrally placed load, the
beam of question 1 has a uniformly distrib¬
uted load of 1 kN/m spread over its span of
length 4 m. Its maximum shearing force is
now:
(a) 4 kN
(c) 2 kN
[see Figure 8.47(a) on page 126]
(b) 1 kN
(d) 0
Bending moment and shearforce diagrams 125
3. A cantilever of length 3 m has a load of 4 kN
placed on its free end. The magnitude of its
maximum bending moment is:
(a) 3 kN m (b) 4 kN m
(c) 12 kN m (d) 4/3 kN/m
4. The magnitude of the maximum shearing
force for the cantilever of question 3 is:
(a) 4 kN (b) 12 kN
(c) 3 kN (d) zero
5. A cantilever of 3 m length carries a UDL of
2 kN/m. The magnitude of the maximum
shearing force is:
(a) 3 kN (b) 2 kN
(c) 6 kN (d) zero
6. In the cantilever of question 5, the magni¬
tude of the maximum bending moment is:
(a) 6 kN m (b) 9 kN m
(c) 2 kN m (d) 3 kN m
Answers to Exercise 51 (page 121)
1.5kNm
BMD
1.5kN
0
1.5kN
-1.5kN
0
-1.5kN
SFD
(a)
1.33kN
1.33kN
-2.67 kN
SFD
-2.67kN
2 kN
0
I
BK/ID
I
I
1 kN
-3 kN
SFD
(c)
-3 kN
SFD
(d)
Figure 8.45
Part Two
Part Two
126 Mechanical Engineering Principles
3.333 kN m
(e)
(f)
3 kN m
( 9 )
Figure 8.45 (Continued)
Answers to Exercise 52 (page 124)
36.75 kN m
90 kN m
(b)
Figure 8.46
A
B
0
-75 kN
0
-30 kN
For fully worked solutions to each of the problems in Practice Exercises 51 to 54 in this chapter,
go to the website:
www.routledge.com/cw/bird
Chapter 9
First and second
moments of area
Why it is important to understand: First and second moment of areas
The first moment of area is usually used to find the centroid or centre of gravity of a two-dimensional
shape or an artefact. The position of the centroid and centre of gravity is fundamental and often it is
very important in very many branches of engineering. The second moment of area can be described as
the geometric ability of a two-dimensional figure or lamina to resist rotation about an axis through its
plane. It is extensively used in structures, where it measures the geometric ability of the structure to
resist bending. Likewise in the hydrostatic stability of ships and yachts, the second moment of area of
a yacht or a ship’s water plane is a measure of the yacht or ship to resist rotations, such as those due to
‘heeling’ and ‘trimming’. This chapter starts by defining the first and second moments of area of two-
dimensional figures and laminas, and extends this to shapes that are useful in engineering, such as the
shapes of the cross-sections of beams. For example, I beams, Tee beams and channel cross-sections are
all very important in structural engineering. First and second moments of force are of much importance
in calculating bending stresses in buildings, bridges, ships, etc.
At the end of this chapter you should be able to:
• define a centroid
• define first moment of area
• calculate centroids using integration
• define second moment of area
• define radius of gyration
• state the parallel axis and perpendicular axis theorems
• calculate the second moment of area and radius of gyration of regular sections using a table of standard results
• calculate the second moment of area of /, T and channel bar beam sections
9.1 Centroids
A lamina is a thin flat sheet having uniform thickness.
The centre of gravity of a lamina is the point where
it balances perfectly, i.e. the lamina’s centre of mo¬
ment of mass. When dealing with an area (i.e. a lam¬
ina of negligible thickness and mass) the term centre
of moment of area or centroid is used for the point
where the centre of gravity of a lamina of that shape
would lie.
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Two
128 Mechanical Engineering Principles
9.2 The first moment of area
The first moment of area is defined as the product of
the area and the perpendicular distance of its centroid
from a given axis in the plane of the area. In Figure 9.1,
the first moment of area A about axis XX is given by
(Ay) cubic units.
Figure 9.1
9.3 Centroid of area between a
curve and the x-axis
Figure 9.2 shows an area PQRS bounded by
the curve y = /(x), the x-axis and ordinates
x = a and x = b. Let this area be divided into
a large number of strips, each of width Sx.
A typical strip is shown shaded drawn at point
(x, y) on /(x). The area of the strip is approximately
rectangular and is given by ySx. The centroid, C,
/ \
has coordinates
/
Figure 9.2
First moment of area of shaded strip about axis
Oy = (ySx)(x) = xySx
Total first moment of area PQRS about axis
. nb
Oy = limit > xySx= xycbc
8x^0^ A a
x—a
(iii)
(iv)
(v)
First moment of area of shaded strip about axis
/ \
Ox = (ySx)
y
v 2 v
1 2
=
Total first moment of area PQRS about axis
„ v . 1 2 1 r b 2 J
Ox= limit > — y =— y dx
Sx->o^2 J 2 Ja J
x=a
r k
Area of PQRS , A = ydx (see Engineering
•J a
Mathematics, 7th Edition, page 551)
Let x and y be the distances of the centroid of
area A about Oy and Ox respectively then:
(x)(^4) = total first moment of area A about axis
r ^
Oy= xy dx
a a
from which, jt =
f xydy
•J a
J ydy
O a
and (y)(A) = total moment of area A about axis
Ox = If y 2 dx
Z a a
from which, J -
r b
I ydx
O a
9.4 Centroid of area between a
curve and the y-axis
If x and y are the distances of the centroid of area
EFGH in Figure 9.3 from Oy and Ox respectively, then,
by similar reasoning as above:
Figure 9.3
y=d
(x)(total area) = limit xSy
\
x
<5v—> 0
y=c
First and second moments of area 129
from which,
i.e. the centroid lies at
/ \
L b
v
2 2
which is at the inter¬
section of the diagonals.
and (y)(total area) = limit
<5y—>0
y-d
J2( x 8y)y =
y—c
from which,
9.5 Worked problems on centroids of
simple shapes
Show, by integration, that the
centroid of a rectangle lies at the intersection of
the diagonals.
Let a rectangle be formed by the line y = b, the x-axis
and ordinates x = 0 and x = L as shown in Figure 9.4. Let
the coordinates of the centroid C of this area be (x 5 y).
Figure 9.4
By integration,
x =
J* xydx J* ( x)(b)cbc
/ ydx f bdx
o Jo
L
X
&x]
bL 2
0 = = L
bL 2
Find the position of the centroid of
the area bounded by the curve y = 3x 2 , the x-axis
and the ordinates x = 0 and x = 2.
If (x, y) are the co-ordinates of the centroid of the
given area then:
Hence the centroid lies at (1.5, 3.6)
Determine by integration the position
of the centroid of the area enclosed by the line
y = 4x, the x-axis and ordinates x = 0 and x = 3.
Let the coordinates of the centroid be (x, y) as shown
in Figure 9.5.
Part Two
Part Two
130 Mechanical Engineering Principles
Then
x =
fl xydy fo xiyAx^jdx
Sl ydx fo
f 4x 2 dx
J o _
T
J o
4x
4 xdx
3
o 36
4 xdx
2x
3
0
18
= 2
- 2
y =
2/0 3y2<fy lfo( 4 x f dx
r 3 ,
18
J y dx
3
3 1
16x 3
/ 1 6x 2 dx 2
Jo
3
o _ 72
18
18
18
= 4
Hence the centroid lies at (2, 4)
In Figure 9.5, ABD is a right-angled triangle. The cen¬
troid lies 4 units from AB and 1 unit from BD showing
that the centroid of a triangle lies at one-third of the
perpendicular height above any side as base.
Now try the following Practice Exercise
Practice Exercise 55 Further problems on
centroids of simple
shapes
In Problems 1 to 5, find the position of the cen¬
troids of the areas bounded by the given curves,
the x-axis and the given ordinates.
1. y — 2x; x = 0, x = 3 [(2, 2)]
2. y = 3x + 2; x = 0, x = 4 [(2.50,4.75)]
3. y = 5x 2 ;x=1,x = 4 [(3.036,24.36)]
4. y = 2x 3 ;x = 0,x = 2 [(1.60,4.57)]
5. y = x(3x+ l);x = -l,x = 0
[(-0.833, 0.633)]
9.6 Further worked problems on
centroids of simple shapes
Determine the co-ordinates of
the centroid of the area lying between the curve
y = 5x- x 2 and the x-axis.
y = 5x-x 2 =x(5 -x). Wheny = 0,x = 0 orx = 5. Hence
the curve cuts the x-axis at 0 and 5 as shown in Figure
9.6. Let the co-ordinates of the centroid be (x, y)
then, by integration,
Figure 9.6
First and second moments of area 131
1
/
2 v
25(125) 6250
\
+ 625
/
125
2.5
Hence the centroid of the area lies at (2.5, 2.5)
(Note from Figure 9.6 that the curve is symmetrical about
x = 2.5 and thus x could have been determined ‘on sight’.)
Locate the centroid of the area en¬
closed by the curve y = 2x 2 , the y -axis and ordi¬
nates + = 1 and = 4, correct to 3 decimal places.
2. Find the coordinates of the centroid of the
area that lies between the curve — = x - 2
x
and the x-axis. [(1, - 0.4)]
3. Determine the coordinates of the centroid of
the area formed between the curve y = 9 - x 2
and the x-axis. [(0, 3.6)]
4. Determine the centroid of the area lying be¬
tween y = 4x 2 , the y-axis and the ordinates
y = 0 and y = 4. " [(0.375,2.40)]
From Section 9.4, *
+*> in*
S‘ x4 ’ /, 4 J+
J.
2
y
4
15
Find the position of the centroid of the area
enclosed by the curve y =V5x, the x-axis
and the ordinate x = 5. [(3.0,1.875)]
Sketch the curve y 2 = 9x between the limits
x = 0 and x = 4. Determine the position of the
centroid of this area. [(2.4, 0)]
i _ 8 _
= 0.568
2 + 2 ;
3^2 J
n 4
4
14
3 V 2
9.7 Second moments of area of
1
r> 4
regular sections
and
y =
r 4
J xdy
14
3+2
/-» 4 y3/2
+ S.
1
j, ^
y 5/2
5
2
14
14
3 V 2
3 V 2
\ (31)
5V2
14
3 V 2
= 2.657
4
The first moment of area about a fixed axis of a lami¬
na of area+(, perpendicular distance y from the centroid
of the lamina is defined as Ay cubic units.
The second moment of area of the same lamina as
above is given by Ay 2 , i.e. the perpendicular distance
from the centroid of the area to the fixed axis is squared.
Second moments of areas are usually denoted by
I and have units of mm 4 , cm 4 , and so on.
Several areas, a { , a 2 , « 3 ,.. at distances y l9 y 2 ,y 3 ,.. from
a fixed axis, may be replaced by a single area A, where
A = a x + a 2 + + .. at distance k from the axis, such
that Ak 2 = ^ay 2
k is called the radius of gyration of area A about the
given axis. Since Ak 2 = j ay 2 = I then the radius of
Hence the position of the centroid is at (0.568, 2.657)
gyration ,k = J^
Now try the following Practice Exercise
Practice Exercise 56 Further problems on
centroids of simple
shapes
1. Determine the position of the centroid of a
sheet of metal formed by the curve y = 4x - x 2
which lies above the x-axis. [(2, 1.6)]
The second moment of area is a quantity much used
in the theory of bending of beams (see Chapter 10),
in the torsion of shafts (see Chapter 12), and in calcu¬
lations involving water planes and centres of pressure
(see Chapter 23).
The procedure to determine the second moment of
area of regular sections about a given axis is (i) to find
the second moment of area of a typical element and (ii)
to sum all such second moments of area by integrating
between appropriate limits.
Part Two
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132 Mechanical Engineering Principles
For example, the second moment of area of the
rectangle shown in Figure 9.7 about axis PP is found
by initially considering an elemental strip of width Sx,
parallel to and distance x from axis PP. Area of shaded
strip = bSx. Second moment of area of the shaded strip
about PP = (x 2 )(bSx).
The second moment of area of the whole rectangle
about PP is obtained by summing all such strips
x—d
between x = 0 and x = d, i.e. x 2 b8x
x=0
6x
P I
Figure 9.7
It is a fundamental theorem of integration that
x=d ~ d
/ (A
x 2 bdx
0
limit
Sx^x x=0
Thus the second moment of area of the rectangle
rd
/ x z dx = b
X 3
J 0
r
U)
l_
id
bd
Since the total area of the rectangle, A = db ,
/ 2 X
then Ip p = (db)
d
3
v
Ad
d
I pp = Ak 2 pp thus k 2 pp =
i.e. the radius of gyration about axis PP ,
1 d 2 d
k pp
Parallel axis theorem
In Figure 9.8, axis GG passes through the centroid C
of area A. Axes DD and GG are in the same plane, are
parallel to each other and distance H apart. The parallel
axis theorem states:
= i GG +AHl
Figure 9.8
Using the parallel axis theorem the second moment
of area of a rectangle about an axis through the cen¬
troid may be determined. In the rectangle shown in
bd 3
Figure 9.9, I pp = -(from above)
d
2
d
2
rr
PT
Ll.
'
-<c
1
A
Sx
P\
Figure 9.9
From the parallel axis theorem
+ (bd)
/
d
2
V
/
i.e.
bd 3 T . bd 3 c ...
-= l rr +-trom which,
3 GG 4
! _ bd 3 bd 3 bd 3
yt i2~
Perpendicular axis theorem
In Figure 9.10, axes OX, OY and OZ are mutually per¬
pendicular. If OX and OY lie in the plane of area^4 then
the perpendicular axis theorem states:
I()Z = I OX + ^OY
First and second moments of area 133
Figure 9.10
A summary of derived standard results for the second
moment of area and radius of gyration of regular sec¬
tions are listed in Table 9.1.
The second moment of area of a hollow cross-
section, such as that of a tube, can be obtained by sub¬
tracting the second moment of area of the hole about its
centroid from the second moment of area of the outer
circumference about its centroid. This is demonstrated
in worked problems 10, 12 and 13 following.
Determine the second moment of
area and the radius of gyration about axes AA, BB
and CC for the rectangle shown in Figure 9.11.
d- 12.0 cm
4
c
1
/
/
/
/
/
/
/
/
\
\
\
\
\
\
\
\
C
b =4.0cm
i
B \ B
\A
Figure 9.11
Table 9.1 Summary of Standard Results of the Second Moments of Areas of Regular Sections
Shape
Position of axis
Second moment of area, I
Radius of gyration, k
Rectangle
length d
(1) Coinciding with b
bd 3
3
d
VT
breadth b
(2) Coinciding with d
db 3
3
b
VT
(3) Through centroid, parallel to b
bd 3
12
d
yj\2
(4) Through centroid, parallel to d
db 3
12
b
Vi 2
Triangle
Perpendicular
height h
(1) Coinciding with b
bh 3
12
h
base b
(2) Through centroid, parallel to base
bh 3
36
h
Vf8
(3) Through vertex, parallel to base
bh 3
4
h
Circle
radius r
diameter d
(1) Through centre perpendicular to
plane (i.e. polar axis)
7tr 4
2
nd 4
or 32
r
(2) Coinciding with diameter
nr 4
4
nd 4
or-
64
r
2
(3) About a tangent
5 nr 4
4
. or 5 ^ 4
64
2
r
Semicircle
radius r
Coinciding with diameter
8
2
Part Two
Part Two
134 Mechanical Engineering Principles
From Table 9.1, the second moment of area about
axis AA ,
Iaa =
bd 3
(4.0)(12.0) 3 .... 4
-—— — = 2304 cm 4
3
Radius of gyration, *AA
d 12.0
Similarly, I BB
db
VT VT
(12.0)(4.0) 3
= 6.93 cm
256 cm 4
and
^BB
4.0
V3
= 2.31 cm
The second moment of area about the centroid of a
bd 2
rectangle is - when the axis through the centroid
1 —
is parallel with the breadth b. In this case, the axis CC
is parallel with the length d
Hence , fg = (12.0)<4.0 ) 3 = 64 cm 4
cc 12
and
12
/ b 4.0
k cc = —^= = —~ = 1.15 cm
Vl2 \jl2
Find the second moment of area
and the radius of gyration about axis PP for the
rectangle shown in Figure 9.12.
40.0 mm
< -►
JJL
,15.0mm
25.0 mm
Figure 9.12
and
_ dh 3
GG ~ ~V2
where d = 40.0 mm
h = 15.0 mm
Hence 1 GG = (40 '° )(li0)3 = 11250 mm 4
GG 12
From the parallel axis theorem,
bp = bG +
where A = 40.0 x 15.0 = 600 mm 2 and
H= 25.0 + 7.5 = 32.5 mm,
the perpendicular distance between GG and PP.
Hence
I pp = 11250 + (600)(32.5) 2 = 645000 mm 4
I pp = Ak 2 pp , from which,
k pp
I
PP -
area
645000
600
= 32.79 mm
Determine the second moment of
area and radius of gyration about axis QQ of the
triangle BCD shown in Figure 9.13.
B
Figure 9.13
Using the parallel axis theorem: Iqq = I GG + AH 2 ,
where 1 GG is the second moment of area about the
centroid of the triangle,
i.e.
= bP = (8.0X12.0) 3
GG 36 36
= 384 cm 4 ,
A is the area of the triangle = —bh
= I (8.0X12.0)
= 48 cm 2
and H is the distance between axes GG and QQ
= 6.0+ - (12.0)
3
= 10 cm
Hence the second moment of area about axis QQ ,
7 ee =384 + (48)(10) 2
= 5184 cm 4
Radius of gyration, k qq =
l /ge -
7
5184
i area \l
48
= 10.4 cm
Determine the second moment of
area and radius of gyration of the circle shown in
Figure 9.14 about axis YY.
First and second moments of area 135
Figure 9.14
In Figure 9.14,
ni n (r >. a A
-= — (2.0 y = 4n cm
4 4
Using the parallel axis theorem,
Iyy ^gg +
where H = 3.0 + 2.0 = 5.0 cm.
Hence, / FF = 47 t + [7t(2 .0) 2 ](5.0) 2
= 47 t + 10Ott = 1047T = 327 cm 4
Radius of gyration,
kyy
I
YY
area
= V26 = 5.10 cm
Determine the second moment of
area of an annular section, about its centroidal axis.
The outer diameter of the annulus is D 1 and its
inner diameter is
Second moment of area of annulus about its centroid,
^xx
= (Ijpf of outer circle about its diameter)
- (I^ of inner circle about its diameter)
nD ? 4 nD, 4
= —r- tt- from Table 9.1
i.e.
Determine the second moment
of area and radius of gyration for the semicircle
shown in Figure 9.15 about axisXY!
Figure 9.15
The centroid of a semicircle lies at
4r
3 K
from its diam¬
eter (see Engineering Mathematics 7th Edition , page
575).
Using the parallel axis theorem: I BB = I GG + AH 2 ,
where (from Table 9.1)
.4 /r(10.0) 4
^BB
nr
8
A =
nr 2 tt(IO.O)
= 3927 mm 4 ,
= 157.1 mm 2
and
H
4.244 mm
Hence, 3927
i.e. 3927
^GG
= 4r = 4(10.0) =
3 n 3 n
= / gg +(157.1)(4.244) 2
= Iq G + 2830, from which,
3927 - 2830 = 1097 mm 4
Using the parallel axis theorem again:
V =/ gg + ^( 15 - 0 + 4 - 244 ) 2
i.e. I xx = 1097 + (157.1)(19.244) 2
= 1097+ 58179 = 59276 mm 4
or 59280 mm 4 , correct to 4 significant figures.
Radius of gyration,
'/
area
l( 59276'
V(TixrJ
= 19.42 mm
Determine the polar second mo¬
ment of area of an annulus about its centre. The
outer diameter of the annulus is D 2 and its inner
diameter is D {
The polar second moment of area is denoted by J.
Hence, for the annulus, J = (J of outer circle about its
centre) - (J of inner circle about its centre)
Part Two
Part Two
136 Mechanical Engineering Principles
i.e.
^ 2 4
32
from Table
9.1
Determine the polar second mo¬
ment of area of the propeller shaft cross-section
shown in Figure 9.16.
Figure 9.16
The polar second moment of area of a circle, J =
nd 4
~32~
The polar second moment of area of the shaded area is
given by the polar second moment of area of the 7.0 cm
diameter circle minus the polar second moment of area
of the 6.0 cm diameter circle.
Hence, from Problem 12, the polar second moment of
area of the cross-section shown
—(7.0 4 - 6.0 4
32 '
-+(1105) = 108.5 cm 4
Determine the second moment of
area and radius of gyration of a rectangular lamina
of length 40 mm and width 15 mm about an axis
through one corner, perpendicular to the plane of
the lamina.
The lamina is shown in Figure 9.17.
Figure 9.17
From the perpendicular axis theorem: l zz = 1^ + lyy
, dtf (40)(15) 3 , cnnn 4
I xx = — = -—— = 45000 mnr
, , bd 3 (15)(40) 3 „ nnnn 4
and * yy = = - 3 -= 320000 mnv
Hence, 7 ZZ = 45000 + 320000
= 365000 mm 4 or 36.5 cm 4
Radius of gyration, k zz
I
zz
365000
area
(40)(15)
= 24.7 mm or 2.47 cm
Determine correct to 3 significant
figures, the second moment of area about axis XX
for the composite area shown in Figure 9.18.
For the semicircle, 1^ —
nr 4 = n(4.0) 4
8 8
- 100.5 cm 4
For the rectangle, ^ xx =
db 3
“T
_ (6.0X8.0) 3
3
= 1024 cm 4
For the triangle, about axis TT through centroid C T ,
( 10 )( 6 . 0) 3
bhi
36
= 60 cm 4
By the parallel axis theorem, the second moment of
area of the triangle about axis XX
n2
-60 +
\ (10X6.0)
8.0 + |( 6 . 0 )
= 3060 cm 4
Total second moment of area about XX = 100.5 + 1024
+ 3060 = 4184.5 = 4180 cm 4 , correct to 3 significant
figures.
First and second moments of area 137
Now try the following Practice Exercise
Practice Exercise 57 Further problems
on second moment
of areas of regular
sections
1. Determine the second moment of area and
radius of gyration for the rectangle shown
in Figure 9.19 about (a) axis AA (b) axis
BB , and (c) axis CC.
(a) 72 cm 4 , 1.73 cm (b) 128 cm 4 , 2.31 cm
(c) 512 cm 4 , 4.62 cm
B
o'
CO
cm
A
<
A
3.0 cm
B
C
Figure 9.19
E E
Figure 9.20
2. Determine the second moment of area and
radius of gyration for the triangle shown in
Figure 9.20 about (a) axis DD (b) axis EE ,
and (c) an axis through the centroid of the
triangle parallel to axis DD.
'(a) 729 cm 4 , 3.67 cm (b) 2187 cm 4 ,'
6.36 cm (c) 243 cm 4 , 2.12 cm
3. For the circle shown in Figure 9.21, find
the second moment of area and radius of
gyration about (a) axis FF and (b) axis HH
H
Figure 9.21
[(a) 201 cm 4 , 2.0 cm (b) 1005 cm 4 ,4.47 cm]
Figure 9.22
4. For the semicircle shown in Figure 9.22,
find the second moment of area and radius
of gyration about axis JJ.
[3927 mm 4 , 5.0 mm]
5. For each of the areas shown in Figure 9.23
determine the second moment of area and
radius of gyration about axis LL , by using
the parallel axis theorem.
Figure 9.23
(a) 335 cm 4 , 4.73 cm (b) 22030 cm 4 ,
14.3 cm (c) 628 cm 4 , 7.07 cm
6. Calculate the radius of gyration of a rectan¬
gular door 2.0 m high by 1.5 m wide about
a vertical axis through its hinge. [0.866 m]
7. A circular door of a boiler is hinged so that
it turns about a tangent. If its diameter is
1.0 m, determine its second moment of
area and radius of gyration about the hinge.
[0.245 m 4 ,0.559 m]
8. A circular cover, centre 0, has a radius of
12.0 cm. A hole of radius 4.0 cm and centre
X , where 0A = 6.0 cm, is cut in the cover.
Determine the second moment of area and
the radius of gyration of the remainder
about a diameter through 0 perpendicular
to OX [14280 cm 4 , 5.96 cm]
9. For the sections shown in Figure 9.24, find
the second moment of area and the radius
of gyration about axis XX.
'(a) 12190 mm 4 , 10.9 mm
(b) 549.5 cm 4 , 4.18 cm
Part Two
Part Two
138 Mechanical Engineering Principles
18.0 mm
3.0 mmrr ~ ~
12.0 mm
X 4.0mm X
(a)
6.0 cm
2.5 err
{2.0 cm
3.0 cm
2.0 cm
x ( b ) x
(a) RSJ (b) Tee beam (c) Channel bar
Figure 9.26 Built-up sections
Figure 9.24
3.0 cm
h—H
16.0 cm
A
9.0 cm
h
4.0 cm
- A
B
4.5 cm
i < » i
9.0 cm
15.0 cm
Calculation of the second moments of area and the posi¬
tion of the centroidal, or neutral, axes for such sections
are demonstrated in the following worked problems.
Determine the second moment of
area about a horizontal axis passing through the
centroid, for the / beam shown in Figure 9.27.
. 0.1 m
n M Thickness = 0.02m
0.2m
> r
Figure 9.27
The centroid of this beam will lie on the horizontal axis
NA, as shown in Figure 9.28.
j 10.0 cm
B (b)
b
*—■C
0.1 m
N -
0.1 m
Figure 9.25
A
e
f
j
k
1
r
1
1
l
V 1
9
h
1
J
V
A
d
10. Determine the second moments of areas
about the given axes for the shapes shown
in Figure 9.25. (In Figure 9.25(b), the cir¬
cular area is removed.)
'I AA = 4224 cm 4 , I BB = 6718 cm 4 ,'
I cc = 37300 cm 4
9.8 Second moment of area for
'built-up'sections
The cross-sections of many beams and members of
a framework are in the forms of rolled steel joists
(RSJ’s or/beams), tees, and channel bars, as shown in
Figure 9.26. These shapes usually afford better bending
resistances than solid rectangular or circular sections.
Figure 9.28
The second moment of area of the / beam is given by:
I NA = (/ of rectangle abdc) - (/ of rectangle efhg )
- (/ of rectangle jkml)
Hence, from Table 9.1,
O.lxO.2 3 0.04 x 0.16 3 0.04 x 0.16 3
IffA ~ 12 12 12
= 6.667 x 10 - 5 - 1.365 x lO ^ 5 - 1.365 x 10 ~ 5
i.e. I NA = 3.937 x 10“ 5 m 4
Determine the second moment
of area about a horizontal axis passing through
the centroid, for the channel section shown in
Figure 9.29.
First and second moments of area 139
h*— 0.1 m-
0.2 m
Thickness = 0.02m
Figure 9.29
The centroid of this beam will be on the horizontal axis
NA, as shown in Figure 9.30.
0.1 m
-<-►
0.1 m
N-
0.1 m
e
f
9
h i
4
Figure 9.30
The second moment of area of the channel section is
given by:
I NA = (/ of rectangle abdc about NA) -
(I of rectangle efhg about NA)
0.1 m
Figure 9.31 Tee beam
In this case, we will first need to find the position of
the centroid, i.e. we need to calculate y in Figure 9.31.
There are several methods of achieving this; the tabular
method is as good as any since it can lead to the use of
a spreadsheet. The method is explained below with the
aid of Table 9.2, below.
First, we divide the tee beam into two rectangles, as
shown in Figure 9.31.
In the first column we refer to each of the two rect¬
angles, namely rectangle (1) and rectangle (2). Thus,
the second row in Table 9.2 refers to rectangle (1) and
the third row to rectangle (2). The fourth row refers
to the summation of each column as appropriate. The
second column refers to the areas of each individual
rectangular element, a.
Thus, area of rectangle (1), a { = 0.1 x 0.02 = 0.002 m 2
and area of rectangle (2), a 2 = 0.18 x 0.02 = 0.0036 m 2
Hence, £ a = 0.002 + 0.0036 = 0.0056 m 2
0.1 x 0 . 2- 5 0.08 x 0.16 3
~ 12 12
- 6.667 xlO- 5 -2.731 x 10 ~ 5
i.e. I NA = 3.936 x 10 -5 m 4
Determine the second moment
of area about a horizontal axis passing through the
centroid, for the tee beam shown in Figure 9.31.
The third column refers to the vertical distance of the
centroid of each individual rectangular element from
the base, namely XX.
Thus, 3 ^ = 0.2 - 0.01 = 0.19 m
i 0.18 nnQ
and y 2 = —— = 0.09 m
In the fourth column, the product ay is obtained
by multiplying the cells of column 2 with the cells of
column 3,
Table 9.2
Column
1
2
3
4
5
6
Row 1
Section
a
y
ay
2
ay A
i
Row 2
a)
0.002
0.19
3.8 x 10^
122 x 10- 5
6.6 x 10- 8
Row 3
( 2 )
0.0036
0.09
3.24 x 1(H
2.916 x 10 - 5
9.72 x 10 - 6
Row 4
s
0.0056
7.04 x 10^
1.014 x 10^
9.786 x 10 - 6
Part Two
Part Two
140 Mechanical Engineering Principles
i.e. a l y l =0.002x0.19 = 3.8 x lfrtn 3
a 2 y 2 = 0.0036 x 0.09 = 3.24 x l(Hm 3
and Y, a y = 3.8 x 10^+ 3.24 x 1(H= 7.04 x l(Hm 3
In the fifth column, the product ay 2 is obtained by mul-
tiplying the cells of column 3 by the cells of column 4,
i.e. ^ ay 2 is part of the second moment of area of the
tee beam about XX ,
i.e. a { y 2 = 0.19 x 3.8 x lO^^^x 10" 5 m 4
a 2 y 2 2 = 0.09 x 3.24 x 10" 4 = 2.916 x 10" 5 m 4
and £ay 2 = 7.22 x 10“ 5 +2.916 x 10- 5
= 1.014 x 10^m 4
In the sixth column, the symbol i refers to the second
moment of area of each individual rectangle about its
own local centroid.
bd 3
Now i = — from Table 9.1
0.1x0.02 3
Hence, i { = -—-= 6.6 x 10 8 m4
0.02 x 0.18 3 , .
L = -—- = 9.72 x 10~ 6 m 4
and Y' = 66 x 10~ 8 + 9.72 x lO^ 6 = 9.789 x lO^ 6 m 4
From the parallel axis theorem: J xx = Y, / + Y, ay 2
(9.1)
The cross-sectional area of the tee beam
= ^ a = 0.0056 m 2 from Table 9.2.
Now the centroidal position, namely y, is given by:
7.04 xlO" 4
0.0056
= 0.1257 m
It should be noted that the least second moment of area
of a section is always about an axis through its centroid.
(a) Determine the second moment
of area and the radius of gyration about axis XX for
the I-section shown in Figure 9.32.
(b) Determine the position of the centroid of the
I-section.
(c) Calculate the second moment of area and
radius of gyration about an axis CC through
the centroid of the section, parallel to axis XX.
The I-section is divided into three rectangles, D , E
and F and their centroids denoted by C D , C E and C F
respectively.
(a) For rectangle D:
The second moment of area about C D (an axis
through C D parallel to XX)
_ _ (8.0)(3.0)> 4
12 12
Using the parallel axis theorem: i X x = 18 + AH 2
where A = (8.0)(3.0) = 24 cm 2 and H= 12.5 cm
Hence 7^= 18 + 24(12.5) 2 = 3768 cm 4
For rectangle E:
From equation (9.1), ^ i + ^ ay 2
= 9.786 xl0" 6 + 1.014 xlO" 4
i.e. ^xx~ 1*112 x 10 _4 m 4
From the parallel axis theorem:
^NA =I XX~ (y) 2 ^L a
= 1.112 x 10- 4 -(0.1257) 2 x 0.0056
I NA = 2.27 x 10 _5 m 4
The second moment of area about C E (an axis
through C E parallel to XX)
bd 3 (3.0)(7.0) 3 4
' IT ’ —JT 2 - ' 85 - 75 cm
Using the parallel axis theorem:
I xx = 85.75 + (7.0)(3.0)(7.5) 2 = 1267 cm 4
For rectangle F\ I xx = = ( 15 -°X 4 - 0 ) 3
= 320 cm 4
3
First and second moments of area 141
Total second moment of area for the I-section
about axis XX ,
4^=3768+ 1267 + 320
= 5355 cm 4
Total area of I-section
= (8.0)(3.0) + (3.0)(7.0) + (15.0)(4.0)
= 105 cm 2
Radius of gyration, k xx =
= 7.14 cm
(b) The centroid of the I-section will lie on the axis
of symmetry, shown as SS in Figure 9.32. Using a
tabular approach:
Table 9.3
Part
Area
( a cm 2 )
Distance of
centroid from
XX (i.e. y cm)
Moment
about XX
(i.e. ay cm 3 )
D
24
12.5
300
E
21
7.5
157.5
F
60
2.0
120
A = 105
^ ay = 577.5
Ay
= ^ay 9 from which,
_ lay 577.5 „
J = +T = 7oT = 5 - 5cm
/ \
5355
area V
+05 J
Now try the following Practice Exercises
Practice Exercise 58 Further problems
on second moment
of area of'built-up'
sections
Determine the second moments of area about a
horizontal axis, passing through the centroids,
for the ‘built-up’ sections shown below. All di¬
mensions are in mm and all the thicknesses are
2 mm.
1 .
Figure 9.33
[17329 mm 4 ]
«-40-►
A
30
V
Thus the centroid is positioned on the axis of
symmetry 5.5 cm from axis XX.
From the parallel axis theorem:
Ixx=Icc + AH2
i.e. 5355 =/ cc +(105)(5.5) :
= / cc +3176
from which, second moment of area about axis
CC, I cc = 5355 -3176
= 2179 cm 4
Radius of gyration, k cc =
/2179
= 4.56 cm
Figure 9.34
3.
[37272 mm 4 ]
Figure 9.35
[18636 mm 4 ]
Part Two
Part Two
142 Mechanical Engineering Principles
4.
Figure 9.36
[10443 mm 4 ]
5.
Figure 9.37
[43909 mm 4 ]
6 .
20
30
Figure 9.38
[8922 mm 4 ]
7.
30
20
Figure 9.39
[3242 mm 4 ]
8 .
20
30
40-
Figure 9.40
[24683 mm 4 ]
Practice Exercise 59 Short-answer
questions on first and
second moment of
areas
1. Define a centroid.
2. Define the first moment of area.
3. Define second moment of area.
4. Define radius of gyration.
5. State the parallel axis theorem.
6. State the perpendicular axis theorem.
Practice Exercise 60 Multiple-choice
questions on first and
second moment of
areas
(Answers on page 335)
1. The centroid of the area bounded by the
curve y = 3x, the x-axis and ordinates x = 0
and x = 3, lies at:
(a) (3, 2) (b) (2, 6)
(c) (2, 3) (d) (6, 2)
2. The second moment of area about axis GG
of the rectangle shown in Figure 9.41 is:
(a) 16 cm 4 (b) 4 cm 4
(c) 36 cm 4 (d) 144 cm 4
60 mm
*** ^ ***
20mm
'
<
X
115mm
X
Figure 9.41
First and second moments of area 143
3. The second moment of area about axis XX
of the rectangle shown in Figure 9.41 is:
(a) 111 cm 4 (b) 31 cm 4
(c) 63 cm 4 (d) 79 cm 4
4. The radius of gyration about axis GG of the
rectangle shown in Figure 9.41 is:
(a) 5.77 mm (b) 17.3 mm
(c) 11.55 mm (d) 34.64 mm
5. The radius of gyration about axis XX of the
rectangle shown in Figure 9.41 is:
(a) 30.41mm (b) 25.66 mm
(c) 16.07 mm (d) 22.91mm
The circumference of a circle is 15.71 mm. Use
this fact in questions 6 to 8.
6. The second moment of area of the circle
about an axis coinciding with its diameter
is:
(a) 490.9 mm 4 (b) 61.36 mm 4
(c) 30.69 mm 4 (d) 981.7 mm 4
7. The second moment of area of the circle
about a tangent is:
(a) 153.4 mm 4 (b) 9.59 mm 4
(c) 2454 mm 4 (d) 19.17 mm 4
8. The polar second moment of area of the
circle is:
(a) 3.84 mm 4 (b) 981.7 mm 4
(c) 61.36 mm 4 (d) 30.68 mm 4
9. The second moment of area about axis XX
of the triangle ABC shown in Figure 9.42
is:
(a) 24 cm 4 (b) 10.67 cm 4
(c) 310.67 cm 4 (d) 324 cm 4
Figure 9.42
10. The radius of gyration about axis GG of the
triangle shown in Figure 9.42 is:
(a) 1.41 cm (b) 2 cm
(c) 2.45 cm (d) 4.24 cm
For fully worked solutions to each of the problems in Practice Exercises 55 to 60 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Two
Part Two
Revision Test 4 Forces in structures, bending moment and shear force diagrams,
and second moments of area
This Revision Test covers the material contained in Chapters 7 to 9. The marks for each question are shown in
brackets at the end of each question.
1. Determine the unknown internal forces in the
pin-jointed truss of Figure RT4.1.
Figure RT4.1
(7)
2. Determine the unknown internal forces in the
pin-jointed truss of Figure RT4.2.
Figure RT4.2
( 12 )
3. A beam of length 3 m is simply supported at its
ends. A clockwise couple of 4 kN m is placed
at a distance of 1 m from the left hand support,
(a) Determine the end reactions, (b) If the beam
now carries an additional downward load of 12 kN
at a distance of 1 m from the right hand support,
sketch the bending moment and shearing force di¬
agrams. (9)
4. A beam of length 4 m is simply supported at its
right extremity and at 1 m from the left extremi¬
ty. If the beam is loaded with a downward load of
2 kN at its left extremity and with another down¬
ward load of 10 kN at a distance of 1 m from its
right extremity, sketch its bending moment and
shearing force diagrams. (7)
5. (a) Find the second moment of area and radius
of gyration about the axis XX for the beam
section shown in Figure RT4.3.
1 60
cm
4
-
1.0 cm
2.0 cm
'
k.
8.0 cm
<
x
i
'
) 2.0 cm
X
s\
-<-
—►
/\
10.0
cm
Figure RT4.3
(b) Determine the position of the centroid of the
section.
(c) Calculate the second moment of area and
radius of gyration about an axis through the
centroid parallel to axis XX. (25)
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 4,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
Chapter 10
Why it is important to understand: Bending of beams
If a beam of symmetrical cross-section is subjected to a bending moment, then stresses due to bending
action will occur. In pure or simple bending, the beam will bend into an arc of a circle. Due to couples, the
upper layers of the beam will be in tension, because their lengths have been increased, and the lower lay¬
ers of the beam will be in compression, because their lengths have been decreased. Somewhere in between
these two layers lies a layer whose length has not changed, so that its stress due to bending is zero. This
layer is called the neutral layer and its intersection with the beam’s cross-section is called the neutral axis.
In this chapter stresses in a beam and the radius of curvature due to bending are calculated. The bend¬
ing of beams is of much importance in designing buildings, bridges, ships, etc., and ensuring their safety
under these loads.
At the end of this chapter you should be able to:
• define neutral layer
• define the neutral axis of a beam’s cross-section
• prove that —
y
M _ E
T~R
• calculate the stresses in a beam due to bending
• calculate the radius of curvature of the neutral layer due to a pure bending moment M
10.1 Introduction
If a beam of symmetrical cross-section is subjected to a
bending moment M, then stresses due to bending action
will occur. This can be illustrated by the horizontal
beam of Figure 10.1, which is of uniform cross-section.
In pure or simple bending, the beam will bend into an
arc of a circle as shown in Figure 10.2.
Figure 10.1
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Two
146 Mechanical Engineering Principles
(a) Beam
(b) Cross-section
Figure 10.2
Now in Figure 10.2, it can be seen that due to these
couples M, the upper layers of the beam will be in ten¬
sion, because their lengths have been increased, and
the lower layers of the beam will be in compression,
because their lengths have been decreased. Some¬
where in between these two layers lies a layer whose
length has not changed, so that its stress due to bend¬
ing is zero. This layer is called the neutral layer and
its intersection with the beam’s cross-section is called
the neutral axis (A£4). Later on in this chapter it will
be shown that the neutral axis is also the centroidal axis
described in Chapter 9.
a M E
In the formula — = —— = —
y I R
o = the stress due to bending moment M occurring at
a distance y from the neutral axis NA,
1= the second moment of area of the beam’s cross-
section about NA,
E = Young’s modulus of elasticity of the beam’s
material, and
R = radius of curvature of the neutral layer of the
beam due to the bending moment M.
Now the original length of the beam element,
dx = R0 (10.1)
At any distance y from NA, the length AB increases its
length to:
CD = (R+ y)0 (10.2)
Hence, extension of AB = d = (R+ y)6 - R0 = y6
Now, strain s = extension/original length,
yO_ = L
R6 R
(10.3)
stress (<7)
However, - ; - =E
strain (e)
or o — Es (10.4)
Substituting equation (10.3) into equation (10.4) gives:
a = E^ (10.5)
a E
or — = — (10.6)
y R
Consider now the stresses in the beam’s cross-section,
as shown in Figure 10.3.
Figure 10.3
From Figure 10.3, it can be seen that the stress o causes
an elemental couple SM about NA, where:
SM = a x (Z? x dy) x y
and the total value of the couple caused by all such
stresses
but from equation (10.5), o
= M= = j *g by dy
Ey
R
(10.7)
Therefore,
Now, E and R are constants, that is, they do not vary
withy, hence they can be removed from under the inte¬
gral sign. Therefore,
However, f y 2 bdy - = I = the second moment
*1 3
of area of the beam’s cross-section about NA (from
Table 9.1, page 133).
Therefore,
or
E T
M_ _ E_
T~r
( 10 . 8 )
Combining equations (10.6) and (10.8) gives:
<7 M_ £
i.e.
(10.9)
Bending of beams 147
Position of NA
From equilibrium considerations, the horizontal force
perpendicular to the beam’s cross-section, due to the
tensile stresses, must equal the horizontal force perpen¬
dicular to the beam’s cross-section, due to the compres¬
sive stresses, as shown in Figure 10.4.
Tensile stress
/
Figure 10.4
Hence,
J* ' <7 bdy = J* ' abdy
or
or
1 <7 bdy - J* ' <7 bdy = 0
f ' G bdy = 0
J -y 2
y
R
But from equation (10.5), o = E
/ y i Ey , ,
-z-bdy = 0
—72 R
Now, E and R are constants, hence
E_
~R
However, Jybdy
= the first moment of area about
the centroid, and where this is zero, coincides with the
centroidal axis, i.e. the neutral axis lies on the same
axis as the centroidal axis.
Moment of resistance ( M)
From Figure 10.4, it can be seen that the system of
tensile and compressive stresses perpendicular to the
beam’s cross-section, cause a couple, which resists the
applied moment M, where
P a ( bd y)y
J -y2
But from equation (10.5), o = E
R
Hence,
M= |
f y 2 bdy = ^
by 3
R J
R
3
or
El
M = — (as required)
R
10.3 Worked problems on the
bending of beams
A solid circular section bar of diam¬
eter 20 mm, is subjected to a pure bending moment
of 0.3 kN m. If E = 2 x 10 n N/m 2 , determine the
resulting radius of curvature of the neutral layer of
this beam and the maximum bending stress.
From Table 9.1, page 133,
nd A nx 20 4
1 =
64
64
= 7854 mm 4
Now,
M= 0.3 kN m x 1000 x 1000^
kN
m
= 3 x 10 5 N mm
and
E = 2x 10
n
N
xl
m
xl
m
m - 1000 mm 1000 mm
= 2 x 10 5 N/mm 2
m_ = e l
T~r
hence, radius of curvature,
From equation (10.8),
* = £= 2xl0 5 ^x 7854mm 4
M
mirr 3xl0 5 Nmm
i.e.
R = 5236 mm = 5.24 m
o M
From equation (10.9), — = —
y i
and
d= Ml
I
where <7 = maximum stress due to bending
~ d 20
and y = outermost fibre from NA = — =
2 2
= 10 mm
Hence, maximum bending stress,
a _ My _ 3 x 10 5 Nmm x 10mm
/ 7854mm 4
= 382 N/mm 2 = 382 x 10 6 N/m 2 = 382 MPa
A beam of length 3 m is simply
supported at its ends and has a cross-section, as
shown in Figure 10.5. If the beam is subjected to a
uniformly distributed load of 2 tonnes/m, deter¬
mine the maximum stress due to bending and the
corresponding value of the radius of curvature of
the neutral layer.
Part Two
Part Two
148 Mechanical Engineering Principles
|
0.2 m
Thickness = 0.02 m
-0.1 m-
Figure 10.5
The total weight on the beam = wL , and as the beam is
symmetrically loaded, the values of the end reactions,
wL
R = as shown in Figure 10.6.
r = ^L
R = ^k
Figure 10.6
Now the maximum bending moment, M, occurs at the
mid-span, where
— L wL L wL
M =R X --—x-bntR = —,
wL L wL L
hence M = x — - — x —
2 2 2 4
wL 2 wL 2
= wl?
Vi'
wL 2
8
tonnes
m
v
8
7
X
( 3m )
8
= 2.25 tonnes m
1000kg 9.81N
= 2.25 tonnes m x-x
tonne
kg
i.e. maximum bending moment, M = 22073 N m
0.1x 0.2 3 0.06 x 0.16 3
Second moment of area, / =-—-—-
= 6.667 x 1(T 5 - 2.048 x 1(T 5
i.e. I = 4.619 x 10" 5 m 4
The maximum stress <7 occurs in the fibre of the
beam’s cross-section, which is the furthest distance
from N A, namely j)
By inspection,
From
<7
A
y
y =
M
T
0.2
= 0.1 m
$ = M_y_ = 22073N m x 0.1m
1 4.619 xl0" 5 m 4
i.e. maximum stress, 6 = 47.79 x 10 6 N/m 2
= 47.79 MPa
A cantilever beam, whose cross-section
is a tube of diameter 0.2 m and wall thickness of
0.02 m, is subjected to a point load, at its free end,
of 3 kN, as shown in Figure 10.7. Determine the
maximum bending stress in this cantilever.
>
3 kN
'
1
* 1 S m w
“ 1 . J III ^
Y/,
Figure 10.7
n ( D i~ D \)
From problem 10, page 135, /= -—-
where D 1 = the external diameter of the tube, and
D { = the internal diameter of the tube.
tt(0.2 4 — 0.16 4 ]
Hence I = —^-rr- L
64
i.e. /= 4.637 xl0- 5 m 4
The maximum bending moment, namely M , will oc¬
cur at the built-in end of the beam, i.e. on the extreme
right of the beam of Figure 10.7.
Maximum bending moment,
M = WxL — 3 kN x 1.5 m x 1000 A
kN
= 4500 N m
The maximum stress occurs at the outermost fibre of
the beam’s cross-section from NA, namely at y
By inspection,
Hence,
0.2
y
= 0.1 m
4500Nm x 0.1m
<7
My _
1 4.637 xl0“ 5 m 4
i.e. the maximum bending stress,
a = 9.70 x 10 6 N/m 2 = 9.70 MPa
A rectangular plank of wood of
length 2.8 m is floating horizontally in still water,
as shown in Figure 10.8. If the cross-section of the
wooden plank is of rectangular form, of width
Bending of beams 149
B = 280 mm and of thickness, D = 80 mm, deter¬
mine the maximum bending stress in the plank,
assuming that a concentrated mass of 320 kg is
placed at its mid span. Let g = 9.81 m/s 2 . Neglect
the self-weight of the plank.
320 kg
)
\ 1 1
< i
K -
k t
k >
V
k i
V
k jk )
k i
k >
k t
k )
k
w
l = 2.8m
Figure 10.8
Let m = mass of ‘weight’ = 320 kg,
and W = weight of the mass = mg
= 320 kg x 9.81 m/s 2 = 3139.2 N
Let w = load per unit length, caused by the water
pressure, acting upwards.
Resolving vertically : upward forces = downward forces
i.e. wi=W
W 3139.2
i.e. w = — = 2 8 - 1121.1 N/m
By inspection, the maximum bending moment, namely
M, will occur at mid-span, where
— I i wi 1
M = w x — x — =-
2 4 8
1121.1 x (2.8)
8
=1099 N m
/ = second moment of area of the rectangular cross-
i_ . , . BD 3
section about its neutral axis =
12
i.e.
1 =
280 x 10" 3 x(80xl0~ 3
12
= 1.195 x 10 5 m 4
y = distance of the fibre of the rectangular cross-
section from the neutral axis, i.e.
D 80 x 10 -3
y = — = --- = 40 x 1(T 3 m
My _ 1099 x 40 x 10" 3
Maximum stress = a = ~T~ ~ --—
/ 1.195 x 10 -5
= 3.68 MPa
Now try the following Practice Exercises
Practice Exercise 61 Further problems on
the bending of beams
1. A cantilever of solid circular cross-section
is subjected to a concentrated load of 30 N
at its free end, as shown in Figure 10.9. If the
diameter of the cantilever is 10 mm, deter¬
mine the maximum stress in the cantilever.
M
Figure 10.9
[367 MPa]
2. If the cantilever of Figure 10.9 were replaced
with a tube of the same external diameter,
but of wall thickness 2 mm, what would be
the maximum stress due to the load shown in
Figure 10.9? [421 MPa]
3. A uniform section beam, simply supported
at its ends, is subjected to a centrally placed
concentrated load of 5 kN. The beam’s
length is 1 m and its cross-section is a solid
circular one.
If the maximum stress in the beam is
limited to 30 MPa, determine the minimum
permissible diameter of the beam’s cross-
section. [75 mm]
4. If the cross-section of the beam of Problem 3
were of rectangular shape, as shown in Fig¬
ure 10.10, determine its dimensions. Bend¬
ing can be assumed to take place about the
xx axis.
D/2
< -►
A
D x
t
D/2
T
X
Figure 10.10
[79.4 mm x 39.7 mm]
Part Two
150 Mechanical Engineering Principles
5. If the cross-section of the beam of
Problem 3 is a circular tube of external diam¬
eter d and internal diameter dl 2, determine
the value of d. [76.8 mm]
6. A cantilever of length 2 m, carries a uni¬
formly distributed load of 30 N/m, as shown
in Figure 10.11. Determine the maximum
stress in the cantilever. [39.1 MPa]
(a) Cantilever
JL
25 mm
T
(b) Cross-section
(solid)
Figure 10.11
7. If the cantilever of Problem 6 were replaced
by a uniform section beam, simply supported
at its ends and carrying the same uniformly
distributed load, determine the maximum
stress in the beam. The cross-section of the
beam may be assumed to be the same as that
of Problem 6. [9.78 MPa]
8. If the load in Problem 7 were replaced by a
single concentrated load of 120 N, placed at
a distance of 0.75 m from the left support,
what would be the maximum stress in the
beam due to this concentrated load?
[36.7 MPa]
9. If the beam of Figure 10.11 were replaced by
another beam of the same length, but which
had a cross-section of tee form, as shown in
Figure 10.12, determine the maximum stress
in the beam.
■< -15 mm-►
Thickness = 5 mm
40 mm
Practice Exercise 62 Short-answer
questions on the
bending of beams
1. Define neutral layer.
2. Define the neutral axis of a beam’s cross-
section.
3. Give another name for the neutral axis.
4. Write down the relationship between stress
o and bending moment M.
5. Write down the relationship between stress
o and radius of curvature R.
Practice Exercise 63 Multiple-choice
questions on the
bending of beams
(Answers on page 335)
1. The maximum stress due to bending occurs:
(a) at the neutral axis
(b) at the outermost fibre
(c) between the neutral axis and the outer¬
most fibre.
2. If the bending moment is increased in a beam,
the radius of curvature will:
(a) increase
(b) decrease
(c) stay the same.
3. If the Young’s modulus is increased in a
beam in bending, due to a constant value of
M, the resulting bending stress will:
(a) increase
(b) decrease
(c) stay the same.
Figure 10.12
[33.9 MPa]
For fully worked solutions to each of the problems in Practice Exercises 61 to 63 in this chapter,
go to the website:
www.routledge.com/cw/bird
Chapter 11
Why it is important to understand: Torque
This chapter commences by defining a couple and a torque. It then shows how the energy and work
done can be calculated from these terms. It then derives the expression which relates torque to the
product of mass moment of inertia and the angular acceleration. The expression for kinetic energy due
to rotation is also derived. These expressions are then used for calculating the power transmitted from
one shaft to another, via a belt. This work is very important for calculating the power transmitted in
rotating shafts and other similar artefacts in many branches of engineering. For example, torque is
important when designing propeller shafts for ships and automobiles, and also for helicopters, etc.
At the end of this chapter you should be able to:
define a couple
define a torque and state its unit
calculate torque given force and radius
calculate work done, given torque and angle turned through
calculate power, given torque and angle turned through
lor
appreciate kinetic energy = —— where / is the moment of inertia
appreciate that torque T— la where a is the angular acceleration
calculate torque given / and a
calculate kinetic energy given / and co
understand power transmission by means of belt and pulley
perform calculations involving torque, power and efficiency of belt drives
11.1 Couple and torque
When two equal forces act on a body as shown in
Figure 11.1, they cause the body to rotate, and the
system of forces is called a couple.
The turning moment of a couple is called a torque, T.
In Figure 11.1, torque = magnitude of either force x
perpendicular distance between the forces,
i.e. T= Fd
d
F
Y
Figure 11.1
The unit of torque is the newton metre, N m
When a force F newtons is applied at a radius r metres
from the axis of, say, a nut to be turned by a spanner,
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Two
152 Mechanical Engineering Principles
as shown in Figure 11.2, the torque T applied to the nut
is given by:
T = Fr N m
! Moment, M
l
Force, F
Turning radius, r
Figure 11.2
Determine the torque when a pulley
wheel of diameter 300 mm has a force of 80 N
applied at the rim.
Torque T = Fr , where force F = 80 N
300
and radius r = - = 150 mm = 0.15 m
2
Hence, torque, T = (80)(0.15) = 12 N m
Determine the force applied
tangentially to a bar of a screw jack at a radius of
800 mm, if the torque required is 600 N m.
Torque, T= force x radius, from which
torque 600 Nm
force = —-— = -r—
radius 800xl0~ J m
= 750 N
The circular hand-wheel of a valve
of diameter 500 mm has a couple applied to it
composed of two forces, each of 250 N. Calculate
the torque produced by the couple.
Torque produced by couple, T = Fd ,
where force F= 250 N
and distance between the forces, d = 500 mm
= 0.5 m
Hence, torque, T= (250)(0.5)
= 125 N m
Now try the following Practice Exercise
Practice Exercise 64 Further problems
on torque
1. Determine the torque developed when a
force of 200 N is applied tangentially to a
spanner at a distance of 350 mm from the
centre of the nut. [70 N m]
2. During a machining test on a lathe, the
tangential force on the tool is 150 N. If
the torque on the lathe spindle is 12 N m,
determine the diameter of the work-piece.
[160 mm]
11.2 Work done and power
transmitted by a constant torque
Figure 11.3(a) shows a pulley wheel of radius r metres
attached to a shaft and a force F newtons applied to the
rim at point P.
Figure 11.3
Figure 11.3(b) shows the pulley wheel having turned
through an angle 6 radians as a result of the force
F being applied. The force moves through a distance s ,
where arc length s = rO
Work done = force x distance moved by the force
= Fxr0 = Fr6 N m = Fr6 J
However, Fr is the torque T, hence,
work done = TO joules
work done _ TO
Average power = time taken ^
stant torque T
However, (angle #)/(time taken) = angular velocity, co
rad/s
for a con-
Hence, power, P = Tea watts (11.1)
Angular velocity, co = Inn rad/s where n is the speed
in rev/s
Hence, power, P = Inn T watts (11.2)
Sometimes power is in units of horsepower (hp),
where 1 horsepower = 745.7 watts
i.e.
1 hp = 745.7 watts
Torque 153
A constant force of 150 N is applied
tangentially to a wheel of diameter 140 mm.
Determine the work done, in joules, in 12 revolutions
of the wheel.
Torque T = Fi% where F = 150 N
140
and radius r = -= 70 mm = 0.070 m
2
Hence, torque T= (150)(0.070) - 10.5 N m
Work done = TO joules, where torque, T= 10.5 N m and
angular displacement,
0= 12 revolutions = 12x2 k rad
= 24k rad.
Hence, work done = TO = (10.5)(24tt) = 792 J
Calculate the torque developed by
a motor whose spindle is rotating at 1000 rev/min
and developing a power of 2.50 kW.
Power P = 2nnT (from above), from which, torque,
P
T= - Nm
2nn
In a turning-tool test, the tangential
cutting force is 50 N. If the mean diameter of the
work-piece is 40 mm, calculate (a) the work done
per revolution of the spindle (b) the power required
when the spindle speed is 300 rev/min.
(a) Work done = TO , where T=Fr
40
Force F = 50 N, radius r = — = 20 mm = 0.02 m
and angular displacement, 0 = 1 rev = 2k rad.
Hence, work done per revolution of spindle
= FrO = (50)(0.02)(2tt) = 6.28 J
(b) Power, P = 2kuT, where
torque, T=Fr = (50)(0.02) = 1 N m and
300
speed ,n
60
= 5 rev/s.
Hence, power required, P = 27t(5)(1) = 31.42 W
A pulley is 600 mm in diameter and
the difference in tensions on the two sides of the
driving belt is 1.5 kN. If the speed of the pulley is
500 rev/min, determine (a) the torque developed,
and (b) the work done in 3 minutes.
where power, P = 2.50 kW = 2500 W
and speed, n = 1000/60 rev/s
Thus, torque,
P
2nn
2500
2k
v
1000
~ 60 ~
/
2500x60
2/rxl000
= 23.87 N m
An electric motor develops a power
of 5 hp and a torque of 12.5 N m. Determine the
speed of rotation of the motor in rev/min.
Power, P = 2kuT, from which,
P
speed n =- rev/s
F 2 kT
where power, P = 5 hp = 5 x 745.7 = 3728.5 W
and torque T= 12.5 N m.
Hence, speed n =
3728.5
2/r(12.5)
= 47.47 rev/s
The speed of rotation of the motor = 47.47 x 60
= 2848 rev/min
(a) Torque T= Fr , where force F= 1.5 kN = 1500 N,
and radius r = ^5. =300 mm = 0.3 m.
2
Hence, torque developed = (1500)(0.3)
= 450 N m
(b) Work done = TO , where torque T= 450 N m
and angular displacement in 3 minutes
= (3 x 500) rev = (3 x 500 x 2k) rad.
Hence, work done = (450)(3 x 500 x 2k)
= 4.24 x 10 6 J
= 4.24 MJ
A motor connected to a shaft
develops a torque of 5 kN m. Determine the
number of revolutions made by the shaft if the
work done is 9 MJ.
Work done = TO , from which, angular displacement,
n= work done
torque
Work done = 9 MJ = 9 x 10 6 J
and torque = 5 kN m = 5000 N m.
Part Two
Part Two
154 Mechanical Engineering Principles
9xio 6
Hence, angular displacement, 9 =
w/ V_/ Y_/ v_/
= 1800 rad.
2k rad = 1 rev, hence, the number of revolutions
1800
made by the shaft = - = 286.5 revs
2 7Z
Now try the following Practice Exercise
Practice Exercise 65 Further problems on
work done and power
transmitted by a
constant torque
1. A constant force of 4 kN is applied tangen¬
tially to the rim of a pulley wheel of diam¬
eter 1.8 m attached to a shaft. Determine the
work done, in joules, in 15 revolutions of the
pulley wheel. [339.3 kJ]
2. A motor connected to a shaft develops a
torque of 3.5 kN m. Determine the number
of revolutions made by the shaft if the work
done is 11.52MJ. [523.8 rev]
3. A wheel is turning with an angular velocity
of 18 rad/s and develops a power of 810 W at
this speed. Determine the torque developed
by the wheel. [45 N m]
4. Calculate the torque provided at the shaft
of an electric motor that develops an out¬
put power of 3.2 hp at 1800 rev/min.
[12.66 Nm]
5. Determine the angular velocity of a shaft
when the power available is 2.75 kW and the
torque is 200 N m. [13.75 rad/s]
6. The drive shaft of a ship supplies a torque of
400 kN m to its propeller at 400 rev/min.
Determine the power delivered by the shaft.
[16.76 MW]
7. A motor is running at 1460 rev/min and
produces a torque of 180 N m. Determine
the average power developed by the motor.
[27.52 kW]
8. A wheel is rotating at 1720 rev/min and
develops a power of 600 W at this speed.
Calculate (a) the torque (b) the work done,
in joules, in a quarter of an hour.
[(a) 3.33 N m (b) 540 kJ]
9. A force of 60 N is applied to a lever of a
scew-jack at a radius of 220 mm. If the
lever makes 25 revolutions, determine
(a) the work done on the jack (b) the power,
if the time taken to complete 25 revolutions
is 40 s. [(a) 2.073 kJ (b) 51.84 W]
11.3 Kinetic energy and
moment of inertia
The tangential velocity v of a particle of mass m mov¬
ing at an angular velocity co rad/s at a radius r metres
(see Figure 11.4) is given by:
v = cor m/s
Figure 11.4
The kinetic energy of a particle of mass m is given by:
1 ?
Kinetic energy = —mv A (from Chapter 16)
= — m(cor) 2 = — m co 2 r 2 joules
^ _
The total kinetic energy of a system of masses rotating
at different radii about a fixed axis but with the same
angular velocity, as shown in Figure 11.5, is given by:
Figure 11.5
Total kinetic energy
1
2
1
— m^co 2 r 2 ^— m 2 co 2 r 2 2
1
+ — m 2 co 2 r 2 2
(m^r 2 + m 2 r 2 2 + m^r 2 )
(O'
2
Torque 155
In general, this may be written as:
Total kinetic energy = (Zmr 2 )
where /(= Yjnr 2 ) is called the moment of inertia of the
system about the axis of rotation and has units of kg m 2 .
The moment of inertia of a system is a measure of the
amount of work done to give the system an angular
velocity of co rad/s, or the amount of work that can be
done by a system turning at co rad/s.
From Section 11.2, work done = TO , and if this work
is available to increase the kinetic energy of a rotating
body of moment of inertia /, then:
TO = I
/ 2 2 A
co 2 -co~
where o^and co 2 a re the initial and
v /
final angular velocities,
/
i.e.
T0 = I
\
C0 2 + CO x
V
{co 2 -co i)
7
/
\
co 2 +ft),
V
is the mean angular velocity,
7
However,
q
i.e. —, where t is the time, and (co 2 - co { ) is the change
t
in angular velocity, i.e. at, where a is the angular
acceleration.
Hence, T0 = I
r \
0
J
(at) from which, torque T = la
where I is the moment of inertia in kg m 2 , a is the an¬
gular acceleration in rad/s 2 and T is the torque in N m.
A shaft system has a moment of inertia
of 37.5 kg m 2 . Determine the torque required to give
it an angular acceleration of 5.0 rad/s 2 .
Torque, T= la, where moment of inertia /= 37.5 kg m 2
and angular acceleration, a = 5.0 rad/s 2 .
Hence, torque, T = Ia = (37.5)(5.0) = 187.5 N m
A shaft has a moment of inertia of
31.4 kg m 2 . What angular acceleration of the shaft
would be produced by an accelerating torque of
495 N m?
Torque, T = la, from which, angular acceleration,
T
a = —, where torque, T = 495 N m and moment of
inertia I = 31.4 kg m 2 .
495
Hence, angular acceleration, a =-
6 31.4
= 15.76 rad/s 2
A body of mass 100 g is fastened to
a wheel and rotates in a circular path of 500 mm in
diameter. Determine the increase in kinetic energy
of the body when the speed of the wheel increases
from 450 rev/min to 750 rev/min.
From above, kinetic energy = I —
Thus, increase in kinetic energy = I
\
\
where moment of inertia, / = mr 2 ,
mass, m = 100 g = 0.1 kg and
radius, r =
500
= 250 mm = 0.25 m.
Initial angular velocity, co l = 450 rev/min
450x2/r
= -—- rad/s
60
= 47.12 rad/s,
and final angular velocity, co 2 = 750 rev/min
750x2;r
60
rad/s
= 78.54 rad/s.
Thus, increase in kinetic energy = I
f 2 2 ^
CO “ - (O x
= (mr 2 )
' 2 2 ^
CO~2~CO x
V
/
(0.1)(0.25 z )
v /
/ 78.54 2 -47.12 2 ^
V
/
= 12.34 J
A system consists of three small
masses rotating at the same speed about the same
fixed axis. The masses and their radii of rotation
are: 15 g at 250 mm, 20 g at 180 mm and 30 g at
200 mm. Determine (a) the moment of inertia of
the system about the given axis, and (b) the kinetic
energy in the system if the speed of rotation is
1200 rev/min.
(a) Moment of inertia of the system, / = X mr 2
i.e. /= [(15 x l(r 3 kg)(0.25 m) 2 ]
+ [(20xl(T 3 kg)(0.18m) 2 ]
+ [(30 x 10~ 3 kg)(0.20 m) 2 ]
= (9.375 x 1(H) + (6.48 x 1(H) + (12 x 1(H)
= 27.855 x l(Hkg m 2 = 2.7855 x 10“ 3 kg m 2
co 2
(b) Kinetic energy = I — , where
moment of inertia, / = 2.7855 x 10 _3 kg m 2
Part Two
Part Two
156 Mechanical Engineering Principles
and angular velocity, co = 2nn = 2k
/ 1200 X
v
60
rad/s
= 40 n rad/s
Hence, kinetic energy in the system
(40 kY
= (2.7855 x 1(T 3 ) V - ’ = 21.99 J
(neglecting any frictional and other resisting
torques).
(b) If the moment of inertia of the rotor is
25 kg m 2 and the speed at the beginning
of the 100 revolutions is 450 rev/min,
determine its speed at the end.
A shaft with its rotating parts has
a moment of inertia of 20 kg m 2 . It is accelerated
from rest by an accelerating torque of 45 N m.
Determine the speed of the shaft in rev/min (a) after
15 s, and (b) after the first 5 revolutions.
Since torque T = la , then angular acceleration,
T 45
a= — = — =2.25 rad/s 2
/ 20
The angular velocity of the shaft is initially zero,
i.e. co { = 0
From Chapter 13, page 173, the angular velocity
after 15 s,
co 2 = co { + at = 0 + (2.25)(15) = 33.75 rad/s,
i.e. speed of shaft after 15 s
/ \
60
(33.75)
2k
rev/min
= 322.3 rev/min
(b) Work done = TO, where torque T = 45 N m and
angular displacement, 0=5 revolutions = 5x2 k
= 107rrad.
Hence work done = (45)(107 t) = 1414 J
This work done results in an increase in kinetic
co 2
energy, given by I — , where moment of inertia
I = 20 kg m 2 and co = angular velocity.
/
Hence, 1414 = (20)
\
co‘
v 2 ,
v
from which,
/ 1414x2 X
co
20
7
= 11.89 rad/s
(a) The kinetic energy gained is equal to the work
done by the accelerating torque of 250 N m over
100 revolutions,
i.e. gain in kinetic energy = work done
= TO = (250)(100 x 2 tt) = 157.08 kJ
(b) Initial kinetic energy is given by:
A=
(25)
/ \2
450x2/r
v
60
7
= 27.76 kJ
The final kinetic energy is the sum of the initial
kinetic energy and the kinetic energy gained,
i.e.
co 1
/— = 27.76 kJ+ 157.08 kJ
= 184.84 kJ
(25) co]
Hence, ' 2
2
184840 from which,
184840x2
25
= 121.6 rad/s
Thus, speed at end of 100 revolutions
121.6x60
= -rev/min = 1161 rev/min
2k
A shaft with its associated rotating
parts has a moment of inertia of 55.4 kg m 2 .
Determine the uniform torque required to accelerate
the shaft from rest to a speed of 1650 rev/min while
it turns through 12 revolutions.
i.e. speed of shaft after the first 5 revolutions
60
= 11.89 x — = 113.5 rev/min
2k
The accelerating torque on a turbine
rotor is 250 N m.
(a) Determine the gain in kinetic energy of the
rotor while it turns through 100 revolutions
From above, T0= I
' 2 2 ^
CO 2 - co~
V
7
i.e.
T= -
I
0
/ 2 2 \
co 2 - co~
V
7
where angular displacement 0= 12 rev = 12 x 27T
= 247T rad,
Torque 157
final speed, co 2 = 1650 rev/min =
1650
60
x 2n
= 172.79 rad/s,
initial speed, co { = 0,
and moment of inertia, 1= 55.4 kg m * 1 2 .
Hence, torque required,
T=
/ \ ( 7 7 A
7 1 co~-co;
V
/
/
55.4 j 172.79 2 -O'
\
V
24;r
-10.97 kN m
7
Now try the following Practice Exercise
Practice Exercise 66 Further problems on
kinetic energy and
moment of inertia
1. A shaft system has a moment of inertia of
51.4 kg m 2 . Determine the torque required to
give it an angular acceleration of 5.3 rad/s 2 .
[272.4 N m]
2. A shaft has an angular acceleration of
20 rad/s 2 and produces an accelerating
torque of 600 N m. Determine the moment
of inertia of the shaft. [30 kg m 2 ]
3. A uniform torque of 3.2 kN m is applied to
a shaft while it turns through 25 revolutions.
Assuming no frictional or other resistances,
calculate the increase in kinetic energy of
the shaft (i.e. the work done). If the shaft is
initially at rest and its moment of inertia is
24.5 kg m 2 , determine its rotational speed,
in rev/min, at the end of the 25 revolutions.
[502.65 kJ, 1934 rev/min]
4. An accelerating torque of 30 N m is applied
to a motor, while it turns through 10 revo¬
lutions. Determine the increase in kinetic
energy. If the moment of inertia of the rotor is
15 kg m2 and its speed at the beginning
of the 10 revolutions is 1200 rev/min,
determine its speed at the end.
[1.885 kJ, 1209.5 rev/min]
5. A shaft with its associated rotating parts has
a moment of inertia of 48 kg m 2 . Determine
the uniform torque required to accelerate the
shaft from rest to a speed of 1500 rev/min
while it turns through 15 revolutions.
[6.283 kN m]
6. A small body, of mass 82 g, is fastened to
a wheel and rotates in a circular path of
456 mm diameter. Calculate the increase in
kinetic energy of the body when the speed
of the wheel increases from 450 rev/min to
950 rev/min. [16.36 J]
7. A system consists of three small masses
rotating at the same speed about the same fixed
axis.The masses and their radii of rotation are:
16 g at 256 mm, 23 g at 192 mm and 31 g at
176 mm. Determine (a) the moment of inertia
of the system about the given axis, and (b) the
kinetic energy in the system if the speed of
rotation is 1250 rev/min.
[(a) 2.857 x 10 3 4 5 kg m 2 (b) 24.48 J]
8. A shaft with its rotating parts has a moment
of inertia of 16.42 kg m 2 . It is accelerated
from rest by an accelerating torque of
43.6 N m. Find the speed of the shaft (a) after
15 s, and (b) after the first four revolutions.
[(a) 380.3 rev/min (b) 110.3 rev/min]
9. The driving torque on a turbine rotor is
203 N m, neglecting frictional and other re¬
sisting torques, (a) What is the gain in kinetic
energy of the rotor while it turns through 100
revolutions? (b) If the moment of inertia of
the rotor is 23.2 kg m 2 and the speed at the be¬
ginning of the 100 revolutions is 600 rev/min,
what will be its speed at the end?
[(a) 127.55 kJ (b) 1167 rev/min]
11.4 Power transmission and
efficiency
A common and simple method of transmitting power
from one shaft to another is by means of a belt passing
over pulley wheels which are keyed to the shafts, as
shown in Figure 11.6. Typical applications include an
electric motor driving a lathe or a drill, and an engine
driving a pump or generator.
Figure 11.6
Part Two
Part Two
158 Mechanical Engineering Principles
For a belt to transmit power between two pulleys there
must be a difference in tensions in the belt on either
side of the driving and driven pulleys. For the direction
of rotation shown in Figure 11.6, F 2 > F {
The torque T available at the driving wheel to do work
is given by:
T=(F 2 -F l )r x Nm
and the available power P is given by:
P= Tco = (F 2 - F 1 )r x w x watts
From Section 11.3, the linear velocity of a point on the
driver wheel, v v = r co Y
A A A
Similarly, the linear velocity of a point on the driven
wheel, v y = r y co y . Assuming no slipping,
v.
v.
y
i.e.
r A
r >’%
Hence
from which,
r x( 27cn x)
fy ( 2 ™y)
r
n
X _
n
Percentage efficiency =
useful work output , ^ ^
energy output x
or
. power output
efficiency = ---— x 100%
power input
An electric motor has an efficiency
of 75% when running at 1450 rev/min. Determine
the output torque when the power input is 3.0 kW.
Efficiency- P ° W "° l '' plll x 100%
power input
hence 75 - P ° W "° ll ' pL " X100
3000
from which, power output
75
x 3000
100
= 2250 W
From Section 11.2, power output, P = 2nnT
P
from which, torque, T
2 nn
where n = (1450/60) rev/s
Hence, output torque =
2250
2 n
1450
~ 60 ~
v
= 14.82 N m
y
A 15 kW motor is driving a shaft at
1150 rev/min by means of pulley wheels and a belt.
The tensions in the belt on each side of the driver
pulley wheel are 400 N and 50 N. The diameters of
the driver and driven pulley wheels are 500 mm and
750 mm respectively. Determine (a) the efficiency of
the motor (b) the speed of the driven pulley wheel.
(a) From above, power output from motor
= (F 2 - F x )r x m x
Force F 2 = 400 N and F { = 50 N,
hence (F 2 -F l ) = 350N,
radius r = = 250 mm = 0.25 m and
x ^
1150x 2/r
angular velocity, co Y = -—- rad/s
60
Hence power output from motor = (F 2 - F { )r x co
/
(350)(0.25)
1150x2;r
v
60
y
= 10.54 kW
Power input = 15 kW
Hence, efficiency of the motor =
- 70.27%
r x _ n y
(b) From above, from which,
v / ’ r n
y x
speed of driven pulley wheels,
n x r x 1150x0.25 y
/?„ = = 767 rev/min
y r 0.750 x
y -
2
A crane lifts a load of mass 5 tonne
to a height of 25 m. If the overall efficiency of the
crane is 65% and the input power to the hauling
motor is 100 kW, determine how long the lifting
operation takes.
The increase in potential energy is the work done and
is given by mgh (see Chapter 16), where mass, m = 5 t
= 5000 kg, g = 9.81 m/s 2 and height h = 25 m.
Hence, work done = mgh = (5000)(9.81)(25)
= 1.226 MJ
Input power = 100 kW = 100000 W
Efficiency - ° l " plllpOW " X100
input power
power output
power input
10 - 54 xl00
15
hence
65 ^ outputpower xlQQ
100000
from which, output power
65
100
x 100000 = 65000 W
work done
time taken
Thus, time taken for lifting operation
workdone _ 1.226 xl0 6 J
output power 65000 W
= 18.86 s
The tool of a shaping machine has
a mean cutting speed of 250 mm/s and the average
cutting force on the tool in a certain shaping
operation is 1.2 kN. If the power input to the motor
driving the machine is 0.75 kW, determine the
overall efficiency of the machine.
Velocity, v = 250 mm/s = 0.25 m/s
and force, F= 1.2 kN = 1200 N
From Chapter 16, power output required at the cutting
tool (i.e. power output),
P = force x velocity = 1200 N x 0.25 m/s
= 300 W
Power input = 0.75 kW = 750 W
Hence, efficiency of the machine
output power
= —-— -x 100
input power
= —x 100 = 40%
750
Calculate the input power of the
motor driving a train at a constant speed of 72 km/h
on a level track, if the efficiency of the motor is
80% and the resistance due to friction is 20 kN.
Force resisting motion = 20 kN = 20000 N and
72
velocity = 72 km/h = — = 20 m/s
3.6
Output power from motor
= resistive force x velocity of train (from Chapter 16)
= 20000 x 20 = 400 kW
Efficiency - p ° werol ' llM x 100
power input
hence 80 =
400
power input
x 100
100
from which, power input = 400 x-= 500 kW
80
Torque 159
Now try the following Practice Exercises
Practice Exercise 67 Further problems on
power transmission
and efficiency
1. A motor has an efficiency of 72% when run¬
ning at 2600 rev/min. If the output torque is
16 N m at this speed, determine the power
supplied to the motor. [6.05 kW]
2. The difference in tensions between the two
sides of a belt round a driver pulley of radius
240 mm is 200 N. If the driver pulley wheel
is on the shaft of an electric motor running at
700 rev/min and the power input to the mo¬
tor is 5 kW, determine the efficiency of the
motor. Determine also the diameter of the
driven pulley wheel if its speed is to be 1200
rev/min. [70.37%, 280 mm]
3. A winch is driven by a 4 kW electric motor
and is lifting a load of 400 kg to a height
of 5.0 m. If the lifting operation takes 8.6 s,
calculate the overall efficiency of the winch
and motor. [57.03%]
4. A belt and pulley system transmits a power
of 5 kW from a driver to a driven shaft.
The driver pulley wheel has a diameter of
200 mm and rotates at 600 rev/min. The
diameter of the driven wheel is 400 mm.
Determine the speed of the driven pulley and
the tension in the slack side of the belt when
the tension in the tight side of the belt is
1.2 kN. [300 rev/min, 404.2 N]
5. The average force on the cutting tool of a lathe
is 750 N and the cutting speed is 400 mm/s.
Determine the power input to the motor
driving the lathe if the overall efficiency
is 55%.
[545.5 W]
6. A ship’s anchor has a mass of 5 tonne.
Determine the work done in raising the an¬
chor from a depth of 100 m. If the hauling
gear is driven by a motor whose output is
80 kW and the efficiency of the haulage is
75%, determine how long the lifting opera¬
tion takes. [4.905 MJ, 1 min 22s]
Part Two
Part Two
160 Mechanical Engineering Principles
Practice Exercise 68 Short-answer
questions on torque
1. In engineering, what is meant by a couple?
2. Define torque.
3. State the unit of torque.
4. State the relationship between work, torque
T and angular displacement 6.
5. State the relationship between power P,
torque T and angular velocity co.
6. Complete the following:
1 horsepower =.watts.
7. Define moment of inertia and state the
symbol used.
8. State the unit of moment of inertia.
9. State the relationship between torque, mo¬
ment of inertia and angular acceleration.
10. State one method of power transmission
commonly used.
11. Define efficiency.
Practice Exercise 69 Multiple-choice
questions on torque
(Answers on page 335)
1. The unit of torque is:
(a) N (b) Pa
(c) N/m (d) N m
2. The unit of work is:
(a) N (b) J
(c) W (d) N/m
3. The unit of power is:
(a) N (b) J
(c) W (d) N/m
4. The unit of the moment of inertia is:
(a) kg m 2 (b) kg
(c) kg/m 2 (d) N m
5. A force of 100 N is applied to the rim of
a pulley wheel of diameter 200 mm. The
torque is:
(a) 2 N m (b) 20 kN m
(c) 10 Nm (d) 20 Nm
6. The work done on a shaft to turn it through
57r radians is 25n J. The torque applied to
the shaft is:
(a) 0.2 Nm (b) 1257r 2 Nm
(c) 307rNm (d) 5Nm
7. A 5 kW electric motor is turning at 50 rad/s.
The torque developed at this speed is:
(a) 100 N m (b) 250Nm
(c) 0.01 Nm (d) 0.1 N m
8. The force applied tangentially to a bar of
a screw-jack at a radius of 500 mm, if the
torque required is 1 kN m is:
(a) 2 N (b) 2 kN
(c) 500 N (d) 0.5 N
9. A 10 kW motor developing a torque of
(200/7 t) N m is running at a speed of:
(a) (7r/20) rev/s (b) 50 n rev/s
(c) 25 rev/s (d) (20/7r) rev/s
10. A shaft and its associated rotating parts has
a moment of inertia of 50 kg m 2 . The angu¬
lar acceleration of the shaft to produce an
accelerating torque of 5 kN m is:
(a) 10 rad/s 2 (b) 250 rad/s 2
(c) 0.01 rad/s 2 (d) 100 rad/s 2
11. A motor has an efficiency of 25%
when running at 3000 rev/min. If the
output torque is 10 N m, the power input is:
(a) 4n kW (b) 0.25 tt kW
(c) \5n kW (d) 75;rkW
12. In a belt-pulley wheel system, the effec¬
tive tension in the belt is 500 N and the
diameter of the driver wheel is 200 mm. If
the power output from the driving motor
is 5 kW, the driver pulley wheel turns at:
(a) 50 rad/s (b) 2500 rad/s
(c) 100 rad/s (d) 0.1 rad/s
For fully worked solutions to each of the problems in Practice Exercises 64 to 69 in this chapter,
go to the website:
www.routledge.com/cw/bird
Chapter 12
Why it is important to understand: Twisting of shafts
The torsion, or twisting, of shafts appears in a number of different branches of engineering, including
propeller shafts for ships and aircraft, shafts driving the blades of a helicopter, shafts driving the rear
wheels of an automobile and shafts driving food mixers, washing machines, tumble dryers, dishwash¬
ers, and so on. If the shaft is overstressed due to a torque, so that the maximum shear stress in the
shaft exceeds the yield shear stress of the shaft’s material, the shaft can fracture. This is an undesirable
phenomenon and normally it should be designed out; hence the need for the theory contained in this
chapter. The twisting of shafts is very important for designing the propulsion systems for ships, auto¬
mobiles, helicopters, and also for food mixers, etc.
At the end of this chapter you should be able to:
• appreciate practical applications where torsion of shafts occurs
T T GO
• prove that ~T~
V J L/
• calculate the shearing stress x, due to a torque, T
• calculate the resulting angle of twist, 0 , due to torque, T
• calculate the power that can be transmitted by a shaft
12.1 To prove that =
t T GO
In the formula — = — = —— :
r J L
t = the shear stress at radius r
T— the applied torque
J = polar second moment of area of the shaft
(note that for non-circular sections, J is
the torsional constant and not the polar
second moment of area)
G = rigidity or shear modulus
0 = angle of twist, in radians, over its
length L
Prior to proving the above formula, the following
assumptions are made for circular section shafts:
(a) the shaft is of circular cross-section
(b) the cross-section of the shaft is uniform along its
entire length
(c) the shaft is straight and not bent
(d) the shaft’s material is homogeneous (i.e. uniform)
and isotropic (i.e. exhibits properties with the
same values when measured along different axes)
and obeys Hooke’s law
(e) the limit of proportionality is not exceeded and
the angles of twist due to the torque are small
(f) plane cross-sections remain plane and normal
during twisting
(g) radial lines across the shaft’s cross-section remain
straight and radial during twisting.
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Two
162 Mechanical Engineering Principles
Consider a circular section shaft, built-in at one end,
namely A, and subjected to a torque T at the other end,
namely B , as shown in Figure 12.1.
Figure 12.1
Let 0 be the angle of twist due to this torque T, where
the direction of T is according to the right hand screw
rule. N.B. The direction of a couple, according to the
right hand screw rule, is obtained by pointing the right
hand in the direction of the double-tailed arrow and
rotating the right hand in a clockwise direction.
From Figure 12.1, it can be seen that:
y = shear strain, and that
y L= RQ (12.1)
provided 0 is small.
However, from equation (3.1), page 57, y =
Hence,
or
/
T
\
L = R6
r _ GO
R~ L
( 12 . 2 )
From equation (12.2), it can be seen that the shear stress
r is dependent on the value of R and it will be a maxi¬
mum on the outer surface of the shaft. On the outer sur¬
face of the shaft r will act as shown in Figure 12.2.
The shaft in Figure 12.2 is said to be in a state of pure
shear on these planes, as these shear stresses will not be
accompanied by direct or normal stress.
Consider an annular element of the shaft, as shown in
Figure 12.3.
L — */
Figure 12.3
The torque T causes constant value shearing stresses on
the thin walled annular element shown in Figure 12.4.
Figure 12.4 Annular element
The elemental torque ST due to these shearing stresses
r at the radius r is given by:
ST = r x (2nrdr)r
and the total torque T= ^ST
or T= f T(2nr 2 )dr (12.4)
Figure 12.2
However, from equation (12.3), r
GO
L
Therefore,
T=
r R °o n 2w
J -j-r{2nr^)dr
But G, 0, L and 2 n do not vary with r,
hence, T= ^-(2 n) J* r 3 dr
L
For any radius r,
T GO
r L
(12.3)
GO kR 4
L 2
Twisting of shafts 163
However, from Table 9.1, page 133, the polar second
moment of area of a circle, J = —-,
2
hence,
GO
L
J
or
T_
7
GO
L
(12.5)
Combining equations (12.3) and (12.5) gives:
r _T__G0
r ~ J ~ L
For a solid section of radius R or diameter D,
n R* nD 4
J = - or
( 12 . 6 )
32
(12.7)
For a hollow tube of circular section and of internal
radius R, and external radius R ~
l
K
j= 2\ R 2- R 1
( 12 . 8 )
or, in terms of external and internal diameters of D 2 and
D { respectively (see Problem 12, Chapter 9, page 135),
(12.9)
The torsional stiffness of the shaft, k , is defined by:
GJ
L
( 12 . 10 )
The next section of worked problems will demonstrate
the use of equation (12.6).
12.2 Worked problems on the
twisting of shafts
An internal combustion engine of
60 horsepower (hp) transmits power to the car
wheels of an automobile at 300 rev/min (rpm).
Neglecting any transmission losses, determine the
minimum permissible diameter of the solid circular
section steel shaft, if the maximum shear stress
in the shaft is limited to 50 MPa. What will be
the resulting angle of twist of the shaft, due to the
applied torque, over a length of 2 m, given that the
rigidity modulus, G = 70 GPa?
(Note that 1 hp = 745.7 W).
Power = 60 hp = 60 hp x 745.7 ^ = 44742 W
hp
From equations (11.1) and (11.2), page 152,
power = Tea = 2 k nT watts,
where n is the speed in rev/s
i.e.
rad 300 rev
44742 = 2k —x—— -x T
rev 60 s
= 31.42 T rad/s
from which,
T=
44742W
31.42 rad/s
i.e. torque T= 1424 N m (since 1 W s = 1 N m)
r _ r
r J
50 x 10 6 N 1424Nm
From equation (12.6),
i.e.
m
nr
and
from which,
2
_ 1424x2
r ;r x 50 xlO 6
3 _ 1424x2
n x 50 x 10 6
nr
or shaft radius, r = 3i
1424x2
n x 50 x 10 6
v
= 0.0263 m
Hence, shaft diameter, d = 2 x r = 2 x 0.0263
= 0.0526 m
T GO
From equation (12.6),
L
N
from which,
0 =
tL
50 x 10 6 —- x 2m
m
G '' 70 x 10 9 A- x 0.0263m
m
and
0 = 0.0543 rad
= 0.0543 rad x
360°
2n rad
i.e. angle of twist, 0 = 3.11*
If the shaft in Problem 1 were
replaced by a hollow tube of the same external
diameter, but of wall thickness 0.005 m, what would
be the maximum shear stress in the shaft due to the
same applied torque, and the resulting twist of the
shaft? The material properties of the shaft may be
assumed to be the same as that of Problem 1.
Internal shaft diameter, D { = D 2 ~ 2 x wall thickness
= 0.0525-2x 0.005
i.e. D x = 0.0425 m
Part Two
Part Two
164 Mechanical Engineering Principles
The polar second moment of area for a hollow tube,
J=
n
32
Hence, J=
D\ - D 4 j from Problem 12, page 135.
K V0525 4 -0.0425 4 )
32
= 4.255 x 10" 7 m 4
r T
From equation (12.6), — = —
r J
Hence, maximum shear stress,
1/10/11VT 0.0525
^ T 1424 N m x-m
T = — =-= 87.85 MPa
J
4.255 xl0“ 7 m 4
T GO
From equation (12.6), — = ——
r L
N
from which,
6 =
tL
87.85 x 10 6 —— x 2m
m
G '' 70 x 10 9 -T x 0.02625m
m
i.e.
0 = 0.0956 rad
= 0.0956 rad x
360°
2k rad
i.e. angle of twist, 0 = 5.48 (
What would be the maximum shear
stress and resulting angle of twist on the shaft
of Problem 2, if the wall thickness were 10 mm,
instead of 5 mm?
Internal shaft diameter, D { = Z) 9 - 2 x wall thickness
= 0.0525-2 x 10 x 1(T 3
i.e.
D x = 0.0325 m
The polar second moment of area for a hollow tube,
71 1 r\ 4
J =
32
Hence, J =
D 2 - j from Problem 12, page 135.
K
0.0525 4 - 0.0325 4
32
= 6.36 x 10 _7 m 4
r T
From equation (12.6), — = —
r J
Hence, maximum shear stress,
,0.0525
- j r 1424 N m x-m
T = __ —
J
- 58.75 MPa
6.36 x 10 -7 m 4
r GO
From equation (12.6), — - ——
r L
from which, 0 =
58.75 xlO 6 —x 2m
GL_ _ _m 2 _
G '' 70 x 10 9 -T x 0.02625m
m
i.e.
0 = 0.06395 rad
= 0.06395 rad x
360
o
In rad
i.e. angle of twist, 0 = 3.66°
N.B. From the calculations in Problems 1 to 3, it can be
seen that a hollow shaft is structurally more efficient
than a solid section shaft.
A shaft of uniform circular section is
fixed at its ends and it is subjected to an intermediate
torque 7, as shown in Figure 12.5, where a > b.
Determine the maximum resulting torque acting on
the shaft and then draw the torque diagram.
c
v B
7"i
A
^ -
^-
T
a
<-
b
-
/
■Tn
Figure 12.5
From equilibrium considerations, (see Figure 12.5),
clockwise torque = the sum of the anticlockwise torques
T=T { + T 2 (12.11)
Let 6 C = the angle of twist at the point C.
T GO
From equation (12.6), — = ——
i/ L/
from which,
Therefore
and
T =
?x =
t 2 =
G0J
L
G0 C J
a
G0 C J
( 12 . 12 )
(12.13)
Dividing equation (12.12) by equation (12.13) gives:
Tx_b
or
T,
a
bTj
a
(12.14)
Substituting equation (12.14) into equation (12.11)
gives:
^ bT 2 rr,
T= —- + T 2 =
a
r \
x + h -
a
T,
\
T
T
or
?2 =
T a
1 +
v
a
/
a + b
v
a
(a + b)
(12.15)
Twisting of shafts 165
However, a + b = L
Therefore
T = —
2 L
(12.16)
Substituting equation (12.16) into equation (12.14) gives:
^ Ta ^
T = = b JL
1 a L
As a > Z?, T 2 > T { ; therefore,
^ Ta
maximum torque = T 2 = -j-
The torque diagram is shown in Figure 12.6.
Tb/L
(12.17)
o
T
Tb/L
-Ta/L
0
-Ta/L
Figure 12.6 Torque diagram
Now try the following Practice Exercises
en that the rigidity modulus, G = 40 GPa?
[76.8 mm, 1.49°]
4. A shaft to the blades of a helicopter trans¬
mits 1000 hp at 200 rev/min. Neglecting
transmission losses, determine the mini¬
mum external and internal diameters of the
hollow circular section aluminium alloy
shaft, when the maximum permissible shear
stress in the shaft is limited to 30 MPa. It
may be assumed that the external diameter
of this shaft is twice its internal diameter.
What will be the resulting angle of twist of
this shaft over a length of 2 m, given that
the modulus of rigidity, G = 26.9 GPa?
[182 mm, 91 mm, 1.5°]
5. A solid circular section shaft of diameter d
is subjected to a torque of 1000 N m. If the
maximum permissible shear stress in this
shaft is limited to 30 MPa, determine the
minimum value of d. [55.4 mm]
Practice Exercise 70 Further problems on
the twisting of shafts
1. A shaft of uniform solid circular section is
subjected to a torque of 1500 N m. Deter¬
mine the maximum shear stress in the shaft
and its resulting angle of twist, if the shaft’s
diameter is 0.06 m, the shaft’s length is 1.2 m,
and the rigidity modulus, G = 77 x 10 9 N/m 2 .
What power can this shaft transmit if it is
rotated at 400 rev/min?
[35.4 MPa, 1.05°, 62.83 kW]
2. If the shaft in Problem 1 were replaced by
a similar hollow one of wall thickness
10 mm, but of the same outer diameter, what
would be the maximum shearing stress in the
shaft and the resulting angle of twist? What
power can this shaft transmit if it is rotated at
500 rev/min? [44.1 MPa, 1.31°, 78.5 kW]
3. A boat’s propeller shaft transmits 50 hp at
100 rev/min. Neglecting transmission losses,
determine the minimum diameter of a solid
circular section phosphor bronze shaft, when
the maximum permissible shear stress in
the shaft is limited to 40 MPa. What will
be the resulting angle of twist of this shaft,
due to this torque, over a length of 1 m, giv-
6. If the shaft in Problem 5 were to be replaced
by a hollow shaft of external diameter d 2 and
internal diameter 0.5 d 2 , determine the mini¬
mum value for d 2 , the design condition being
the same for both shafts. What percentage
saving in weight will result by replacing the
solid shaft by the hollow one?
[56.6 mm, 28.3 mm, 21.7%]
7. An internal combustion engine of a ship is
of power 3000 hp and it transmits its power
through two shafts at 300 rev/min (rpm). If
the transmission losses are 10%, determine
the minimum possible diameters of the
shafts, given that the maximum permissible
shear stress in the shafts is 80 MPa. (It may
be assumed that 1 hp = 745.7 W).
[d = 159 mm]
Practice Exercise 71 Short-answer
questions on the
twisting of shafts
1. State three practical examples where the
torsion of shafts appears.
2. State the relationship between shear stress
r and torque T for a shaft.
Part Two
Part Two
166 Mechanical Engineering Principles
3. State the relationship between torque 7 and
angle of twist 6 for a shaft.
4. State whether a solid shaft or a hollow shaft
is structurally more efficient.
Practice Exercise 72 Multiple-choice
questions on the
twisting of shafts
(Answers on page 335)
1. The maximum shear stress for a solid shaft
occurs:
(a) at the outer surface
(b) at the centre
(c) in between the outer surface and the
centre
2. For a given shaft, if the values of torque 7,
length L and radius r are kept constant, but
rigidity G is increased, the value of shear
stress r:
(a) increases
(b) stays the same
(c) decreases
3. If for a certain shaft, its length is doubled,
the angle of twist:
(a) doubles (b) halves
(c) remains the same
4. If a solid shaft is replaced by a hollow shaft
of the same external diameter, its angle of
twist:
(a) decreases (b) stays the same
(c) increases
5. If a shaft is fixed at its two ends and subject¬
ed to an intermediate torque 7 at mid-length,
the maximum resulting torque is equal to:
(a) T (b) 4
(c) zero
6. If a hollow shaft is subjected to a torque 7,
the shear stress on the inside surface is:
(a) a minimum (b) a maximum
(c) zero
References
[1] A link to the Twisting of Circular Section Shafts -
www.routledge.com/cw/bird
A video reference on YouTube which demonstrates a stan¬
dard experimental test on the twisting of circular cross-
section shafts. This experiment has to be carried out to
determine the torsional material properties of a metal or a
similar material.
For fully worked solutions to each of the problems in Practice Exercises 70 to 72 in this chapter,
go to the website:
www.routledge.com/cw/bird
Revision Test 5 Bending of beams, torque and twisting of shafts
This Revision Test covers the material contained in Chapters 10 to 12. The marks for each question are shown in
brackets at the end of each question.
1. A beam simply supported at its ends, is of length
1.4 m. If the beam carries a centrally-placed down¬
ward concentrated load of 50 kN, determine the
minimum permissible dimensions of the beam’s
cross-section, given that the maximum permis¬
sible stress is 40 MPa, and the beam has a solid
circular cross-section. (6)
2. Determine the force applied tangentially to a bar
of a screw-jack at a radius of 60 cm, if the torque
required is 750 N m. (3)
3. Calculate the torque developed by a motor whose
spindle is rotating at 900 rev/min and developing a
power of 4.20 kW. (5)
4. A motor connected to a shaft develops a torque
of 8 kN m. Determine the number of revolutions
made by the shaft if the work done is 7.2 MJ. (6)
5. Determine the angular acceleration of a shaft that
has a moment of inertia of 32 kg m 2 produced by
an accelerating torque of 600 N m. (5)
6. An electric motor has an efficiency of 72% when
running at 1400 rev/min. Determine the output
torque when the power input is 2.50 kW. (5)
7. A solid circular section shaft is required to trans¬
mit 60 hp at 1000 rpm. If the maximum permis¬
sible shear stress in the shaft is 35 MPa, determine
the minimum permissible diameter of the shaft.
Determine the resulting angle of twist of the shaft
per metre, assuming that the modulus of rigidity,
G = 70 GPa and 1 hp = 745.7 W (10)
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 5,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
Part Two
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Part Three
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Chapter 13
Why it is important to understand: Linear and angular motion
This chapter commences by defining linear and angular velocity and also linear and angular acceleration.
It then derives the well-known relationships, under uniform acceleration, for displacement, velocity and
acceleration, in terms of time and other parameters. The chapter then uses elementary vector analysis,
similar to that used for forces in Chapter 7, to determine relative velocities. This chapter deals with the
basics of kinematics. A study of linear and angular motion is important for the design of moving vehicles.
At the end of this chapter you should be able to:
• appreciate that 2k radians corresponds to 360 °
• define linear and angular velocity
• perform calculations on linear and angular velocity using co = 2nn and v = (Or
• define linear and angular acceleration
• perform calculations on linear and angular acceleration using co 2 = (Q 1 + cct and a = ra
• select appropriate equations of motion when performing simple calculations
• appreciate the difference between scalar and vector quantities
• use vectors to determine relative velocities, by drawing and by calculation
13.1 The radian
The unit of angular displacement is the radi¬
an, where one radian is the angle subtended at the cen¬
tre of a circle by an arc equal in length to the radius,
as shown in Figure 13.1.
The relationship between angle in radians ( 6 ), arc
length ( 5 ) and radius of a circle (r) is:
s = rO (13.1)
Since the arc length of a complete circle is 2nr and the
angle subtended at the centre is 360°, then from equa¬
tion (13.1), for a complete circle,
2 nr = rQ or 0 = 2n radians
Thus, 2n radians corresponds to 360° (13.2)
as stated in Chapter 1.
13.2 Linear and angular velocity
Figure 13.1
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Linear velocity v is defined as the rate of change of
linear displacement s with respect to time t , and for mo¬
tion in a straight line:
Part Three
172 Mechanical Engineering Principles
linear velocity
change of displacement
change of time
5
v — —
t
i.e. v — — (13.3)
The unit of linear velocity is metres per second (m/s).
Angular velocity
The speed of revolution of a wheel or a shaft is usually
measured in revolutions per minute or revolutions
per second, but these units do not form part of a
coherent system of units. The basis used in SI units is
the angle turned through (in radians) in one second.
Angular velocity is defined as the rate of change of
angular displacement 6 , with respect to time t, and for
an object rotating about a fixed axis at a constant speed:
, . angle turned through
angular velocity =-:---
time taken
0
i.e. co = ~ (13.4)
*
The unit of angular velocity is radians per second (rad/s).
An object rotating at a constant speed of n revolutions
per second subtends an angle of 2/m radians in one sec¬
ond, that is, its angular velocity,
co = Inn rad/s (13.5)
From equation (13.1), s = rO , and from equation (13.4),
0 = cot, hence
s = rcot or - = cor
t
s
However, from equation (13.3), v = -
t
hence v = cor (13.6)
Equation (13.6) gives the relationship between linear
velocity, v, and angular velocity, co.
A wheel of diameter 540 mm is
rotating at (1500//r) rev/min. Calculate the angular
velocity of the wheel and the linear velocity of a
point on the rim of the wheel.
From equation (13.5), angular velocity co = 2nn, where
n is the speed of revolution in revolutions per second,
i.e. n
1500
60 /r
revolutions per second.
/
Thus, angular velocity, co = In
1500
60n
\
= 50 rad/s
The linear velocity of a point on the rim, v = cor , where
r is the radius of the wheel, i.e.
r— 0.54/2 or 0.27 m.
Thus, linear velocity, v= cor = 50 x 0.27 = 13.5 m/s
A car is travelling at 64.8 km/h and
has wheels of diameter 600 mm.
(a) Find the angular velocity of the wheels in
both rad/s and rev/min.
(b) If the speed remains constant for 1.44 km,
determine the number of revolutions made by
a wheel, assuming no slipping occurs.
.. o i /, o km i aaa m 1 h
64.8 km/h = 64.8 -j— x 1000 -— x —
h km 3600 s
64.8
= X6 wJs
= 18 m/s
i.e. the linear velocity, v, is 18 m/s
The radius of a wheel is (600/2) mm = 0.3 m
From equation (13.6), v = cor, hence co= v/r
18
i.e. the angular velocity, co = — = 60 rad/s
From equation (13.5), angular velocity, co = Inn,
where n is in revolutions per second. Hence
n = colln and angular speed of a wheel in revolu¬
tions per minute is 60ft)/2/r; but co = 60 rad/s,
60x60
hence angular speed = —-- = 573 revolu-
jLJI
tions per minute (rpm)
From equation (13.3), time taken to travel
1.44 km at a constant speed of 18 m/s is:
1440 m
18 m/s
= 80 s.
Since a wheel is rotating at 573 revolutions per
minute, then in 80/60 minutes it makes
573 x 80
60
= 764 revolutions.
Now try the following Practice Exercise
Practice Exercise 73 Further problems on
linear and angular
velocity
1. A pulley driving a belt has a diameter of
360 mm and is turning at 2100/n revolutions
per minute. Find the angular velocity of the
pulley and the linear velocity of the belt
assuming that no slip occurs.
[co = 90 rad/s, v = 16.2 m/s]
2. A bicycle is travelling at 36 km/h and the
diameter of the wheels of the bicycle is
Linear and angular motion 173
500 mm. Determine the angular velocity of
the wheels of the bicycle and the linear veloc¬
ity of a point on the rim of one of the wheels.
[co = 40 rad/s, v = 10 m/s]
13.3 Linear and angular acceleration
Linear acceleration, a , is defined as the rate of change
of linear velocity with respect to time. For an object
whose linear velocity is increasing uniformly:
1 change of linear velocity
linear acceleration =-:--
time taken
t
The unit of linear acceleration is metres per second
squared (m/s 2 ). Rewriting equation (13.7) with v 2 as
the subject of the formula gives:
v 2 = v x +at (13.8)
where v 2 = final velocity and Vj = initial velocity.
Angular acceleration, a, is defined as the rate of
change of angular velocity with respect to time. For an
object whose angular velocity is increasing uniformly:
1 1 . change of angular velocity
angular acceleration =-:--
time taken
(13.7)
i.e.
a =
(13.9)
The unit of angular acceleration is radians per second
squared (rad/s 2 ). Rewriting equation (13.9) with co 2 as
the subject of the formula gives:
(o 2 = ( o l + at (13.10)
The speed of a shaft increases
uniformly from 300 revolutions per minute to
800 revolutions per minute in 10 s. Find the angular
acceleration, correct to 3 significant figures.
co 2 - co.
From equation (13.9), a= -
Initial angular velocity,
cOj = 300 rev/min = 300/60 rev/s
300 x 2n
60
rad/s,
final angular velocity,
800 x In
co-
60
rad/s and time, t— 10 s.
Hence, angular acceleration,
800 x 2 n 300 x 2 n
a =
—- rad/s 2
500 x 2 n .
——77T = 5 - 24 rad/s 2
60 x 10
If the diameter of the shaft in
problem 3 is 50 mm, determine the linear
acceleration of the shaft on its external surface,
correct to 3 significant figures.
From equation (13.11), a = ra
50
The shaft radius is — mm = 25 mm = 0.025 m, and the
angular acceleration, a = 5.24 rad/s 2 ,
thus the linear acceleration,
a = ra= 0.025 x 5.24 = 0.131 m/s 2
Now try the following Practice Exercise
where co 2 = final angular velocity and C 0 j = initial
angular velocity.
From equation (13.6), v = cor. For motion in a circle
having a constant radius r, v 2 = co 2 r and = coj/%
hence equation (13.7) can be rewritten as:
co 2 r - co { r _ r[co 2 - co { )
co 2 - co.
But from equation (13.9), - = a
Practice Exercise 74 Further problems on
linear and angular
acceleration
1. A flywheel rotating with an angular velocity
of 200 rad/s is uniformly accelerated at a rate
of 5 rad/s 2 for 15 s. Find the final angular
velocity of the flywheel both in rad/s and
revolutions per minute.
[275 rad/s, 825 01n rev/min]
Hence
a = ra
(13.11)
Part Three
Part Three
174 Mechanical Engineering Principles
2. A disc accelerates uniformly from 300
revolutions per minute to 600 revolutions
per minute in 25 s. Determine its angular
acceleration and the linear acceleration of a
point on the rim of the disc, if the radius of
the disc is 250 mm.
[0.4tt rad/s 2 , 0.17F m/s 2 ]
13.4 Further equations of motion
From equation (13.3), s = vt , and if the linear velocity is
changing uniformly from v x to v 2 , then s = mean linear
velocity x time
i.e.
/
5 =
\
vi+v 2
V
t
y
(13.12)
From equation (13.4), 6 = cot , and if the angular
velocity is changing uniformly from co { to co 2 then
6 = mean angular velocity x time
i.e.
/
co 1 + co 2
V
2
(13.13)
Two further equations of linear motion may be derived
from equations (13.8) and (13.11):
s = v l t+\at 1 (13.14)
and v 2 = v 2 + las (13.15)
Two further equations of angular motion may be
derived from equations (13.10) and (13.13):
0 = co x t + ~ at 1 (13.16)
and co 2 2 = co 2 + 2«0 (13.17)
Table 13.1, on page 175, summarises the principal
equations of linear and angular motion for uniform
changes in velocities and constant accelerations and
also gives the relationships between linear and angular
quantities.
The speed of a shaft increases uni¬
formly from 300 rev/min to 800 rev/min in 10 s.
Find the number of revolutions made by the shaft
during the 10 s it is accelerating.
From equation (13.13), angle turned through,
0 =
c \
co x + co 2
t
y
300 x 2 n 800 x 2 n
- + -
\
60
60
v
(10) rad
y
However, there are 2 n radians in 1 revolution, hence,
number of revolutions
X 300x2 k 800x 2
- + -
60
60
r \o ^
K 2n J
y
]_
2
c \
1100
v
60
( 10 )
y
1100
~Y2
= 91.67 revolutions
The shaft of an electric motor,
initially at rest, accelerates uniformly for 0.4 s at
15 rad/s 2 . Determine the angle (in radians) turned
through by the shaft in this time.
1 9
From equation (13.16), 6 = co^ + — at 1
Since the shaft is initially at rest, cOj = 0 and
6 = — at 2 , the angular acceleration, a — 15 rad/s 2 and
time t = 0.4 s.
Hence, angle turned through,
1 9
6= 0 + -x 15 x 0.4 2 = 1.2 rad
A flywheel accelerates uniformly at
2.05 rad/s 2 until it is rotating at 1500 rev/min. If
it completes 5 revolutions during the time it is
accelerating, determine its initial angular velocity
in rad/s, correct to 4 significant figures.
Since the final angular velocity is 1500 rev/min,
rev lmin In rad
co ? = 1500 —r-x—— x —-
z mm 60s lrev
= 50/r rad/s
Linear and angular motion 175
Table 13.1
s = arc length (m)
t = time (s)
v = linear velocity (m/s)
v 1 = initial linear velocity (m/s)
v 2 = final linear velocity (m/s)
a = linear acceleration (m/s 2 )
n = speed of revolution (rev/s)
Equation number
(13.1)
(13.2)
(13.3) and (13.4)
(13.5)
(13.6)
(13.7) and (13.9)
(13.8) and (13.10)
(13.11)
(13.12) and (13.13)
(13.14) and (13.16)
(13.15) and (13.17)
Linear motion
v
a
v 2" v l
v 7 = (vj + at) m/s
s =
s = v l t+ at
v 2 = v \ + 2 as
r
6
CO
CO
CO
a
l
radius of circle (m)
angle (rad)
angular velocity (rad/s)
initial angular velocity (rad/s)
final angular velocity (rad/s)
angular acceleration (rad/s 2 )
Angular motion
s = r6 m
2n rad = 360°
0
co= — rad/s
t
co = Inn rad/s
v = cor m/s 2
a
co 2 ~co i
co ? = ( C 0 j + at) rad/s
a = r a
m/s 2
/
\
\
Vj+V 2
t 9 =
co x + co 2
2
?
y
V
7
6= co 1 t+ at 2
co 2 = co 2 + 2 aO
5 revolutions = 5 rev x- = 10/r rad
lrev
From equation (13.17), co\ = co 2 + 2a6
i.e. (507t) 2 = co 2 + (2 x 2.05 x 10/r)
from which, co 2 = (50zr) 2 - (2 x 2.05 x 10/r)
= (50/r) 2 - 41 tt= 24545
i.e. V24545 = 156.7 rad/s
Thus, the initial angular velocity is 156.7 rad/s,
correct to 4 significant figures.
Now try the following Practice Exercise
Practice Exercise 75 Further problems on
equations of motion
1. A grinding wheel makes 300 revolu¬
tions when slowing down uniformly from
1000 rad/s to 400 rad/s. Find the time for this
reduction in speed. [2.693 s]
2. Find the angular retardation for the grinding
wheel in question 1. [222.8 rad/s 2 ]
Part Three
Part Three
176 Mechanical Engineering Principles
3. A disc accelerates uniformly from 300
revolutions per minute to 600 revolutions
per minute in 25 s. Calculate the number
of revolutions the disc makes during this
accelerating period. [ 187.5 revolutions]
4. A pulley is accelerated uniformly from rest
at a rate of 8 rad/s 2 . After 20 s the accelera¬
tion stops and the pulley runs at constant
speed for 2 min, and then the pulley comes
uniformly to rest after a further 40 s. Calcu¬
late:
(a) the angular velocity after the period of
acceleration
(b) the deceleration
(c) the total number of revolutions made by
the pulley.
[(a) 160 rad/s (b) 4 rad/s 2 (c) 12000//rrev]
13.5 Relative velocity
Quantities used in engineering and science can be
divided into two groups:
(a) Scalar quantities have a size or magnitude only
and need no other information to specify them.
Thus 20 centimetres, 5 seconds, 3 litres and
4 kilograms are all examples of scalar quantities.
(b) Vector quantities have both a size (or magnitude),
and a direction, called the line of action of the
quantity. Typical vector quantities include velocity,
acceleration and force. Thus, a velocity of 30 km/h
due west, and an acceleration of 7 m/s 2 acting
vertically downwards, are both vector quantities.
A vector quantity is represented by a straight line
lying along the line of action of the quantity, and having
a length that is proportional to the size of the quantity, as
shown in Chapter 5. Thus ab in Figure 13.2 represents
a velocity of 20 m/s, whose line of action is due west.
0 5 10 15 20
Scale : velocity in m/s
S
- < -
b a
The bold letters, ab , indicate a vector quantity and the
order of the letters indicate that the line of action is
from a to b.
Consider two aircraft A and B flying at a constant
altitude, A travelling due north at 200 m/s and B trav¬
elling 30° east of north, written N 30° E, at 300 m/s,
as shown in Figure 13.3.
0 100 200 300
Scale: velocity in m/s
Figure 13.3
Relative to a fixed point 0, 0a represents the velocity
of A and 0b the velocity of B. The velocity of B
relative to A, that is the velocity at which B seems to
be travelling to an observer on A, is given by ab , and
by measurement is 160 m/s in a direction E 22 °N. The
velocity of A relative to B , that is, the velocity at which
A seems to be travelling to an observer on B , is given
by ba and by measurement is 160 m/s in a direction
W 22° S.
Two cars are travelling on horizontal
roads in straight lines, car A at 70 km/h at N 10 ° E
and car B at 50 km/h at W 60 °N. Determine, by
drawing a vector diagram to scale, the velocity of
car A relative to car B.
With reference to Figure 13.4(a), oa represents the
velocity of car A relative to a fixed point o , and ob
represents the velocity of car B relative to a fixed
point o. The velocity of car A relative to car B is given
by ba and by measurement is 45 km/h in a direction of
E 35° N.
25
N
W
Figure 13.2
Linear and angular motion 177
0 20 40 60
Scale : velocity in km/h
Figure 13.4
Verify the result obtained in Problem
8 by calculation.
The triangle shown in Figure 13.4(b) is similar to the
vector diagram shown in Figure 13.4(a). Angle BOA is
40°. Using the cosine rule (see Chapter 1):
BA 2 = 50 2 + 70 2 - 2 x 50 x 70 x cos 40°
from which, BA = 45.14
Using the sine rule:
50 45.14
———' - ^ ~ - (also from Chapter 1)
sin ZB AO sin 40° F
50 sin 40°
from which, sin ZBAO= — . _ 1 . — =0.7120
45.14
Hence, angle BAO = 45.40°
thus, angle ABO = 180 °- (40 ° + 45.40 °)
= 94.60°,
and angle 6 = 94.60°- 60°= 34.60°
Thus, ba is 45.14 km/h in a direction E 34.60° N by
calculation.
A crane is moving in a straight line
with a constant horizontal velocity of 2 m/s. At the
same time it is lifting a load at a vertical velocity of
5 m/s. Calculate the velocity of the load relative to
a fixed point on the Earth’s surface.
A vector diagram depicting the motion of the crane
and load is shown in Figure 13.5. oa represents the
velocity of the crane relative to a fixed point on the
Earth’s surface and ab represents the velocity of
the load relative to the crane. The velocity of the load
relative to the fixed point on the Earth’s surface is ob.
By Pythagoras’ theorem (from Chapter 1):
Figure 13.5
ob 2 = oa 2 + ab 2
= 4 + 25 = 29
Hence ob = ^29 = 5.385 m/s
5
Tan 6= — = 2.5, hence, 0 = tan * 1 2.5 = 68.20°
i.e. the velocity of the load relative to a fixed point on
the Earth’s surface is 5.385 m/s in a direction 68.20°
to the motion of the crane.
Now try the following Practice Exercises
Practice Exercise 76 Further problems on
relative velocity
1. A car is moving along a straight horizontal
road at 79.2 km/h and rain is falling verti¬
cally downwards at 26.4 km/h. Find the
velocity of the rain relative to the driver of
the car.
[83.5 km/h at 71.6°to the vertical]
2. Calculate the time needed to swim across
a river 142 m wide when the swimmer can
swim at 2 km/h in still water and the river is
flowing at 1 km/h. At what angle to the bank
should the swimmer swim?
[4 min 55 s, 60 °\
Part Three
Part Three
178 Mechanical Engineering Principles
3. A ship is heading in a direction N 60 ° E at a
speed which in still water would be 20 km/h.
It is carried off course by a current of 8 km/h
in a direction of E 50 ° S. Calculate the ship’s
actual speed and direction.
[22.79 km/h in a direction E 9.78 °N]
Practice Exercise 77 Short-answer
questions on linear
and angular motion
1. State and define the unit of angular
displacement.
2. Write down the formula connecting an
angle, arc length and the radius of a circle.
3. Define linear velocity and state its unit.
4. Define angular velocity and state its unit.
5. Write down a formula connecting angular
velocity and revolutions per second in
coherent units.
6 . State the formula connecting linear and
angular velocity.
7. Define linear acceleration and state its unit.
8 . Define angular acceleration and state its
unit.
9. Write down the formula connecting linear
and angular acceleration.
10. Define a scalar quantity and give two
examples.
11. Define a vector quantity and give two
examples.
Practice Exercise 78 Multiple-choice
questions on linear
and angular motion
(Answers on page 335)
1. Angular displacement is measured in:
(a) degrees (b) radians
(c) rev/s (d) metres
An angle of
3 n
T
radians is equivalent to:
(a) 270°
(c) 135°
3.
5.
An angle of 120 ° is equivalent to:
(b) - rad
3
(d) I rad
An angle of 2 rad at the centre of a circle
subtends an arc length of 40 mm at the
circumference of the circle. The radius of
the circle is:
(a) 40/rmm (b) 80 mm
(c) 20 mm (d) (40/ n) mm
A point on a wheel has a constant angu¬
lar velocity of 3 rad/s. The angle turned
through in 15 seconds is:
(a) 45 rad (b) 10/r rad
(c) 5 rad (d) 90;r rad
An angular velocity of 60 revolutions per
minute is the same as:
(a) (1/27T) rad/s (b) 120;rrad/s
(c) (30/7T) rad/s (d) 2n rad/s
7. A wheel of radius 15 mm has an angular
velocity of 10 rad/s. A point on the rim of
the wheel has a linear velocity of:
(a) 300/rmm/s (b) 2/3 mm/s
(c) 150 mm/s (d) 1.5 mm/s
8 . The shaft of an electric motor is rotat¬
ing at 20 rad/s and its speed is increased
uniformly to 40 rad/s in 5 s. The angular
acceleration of the shaft is:
(a) 4000 rad/s 2 (b) 4 rad/s 2
(c) 160rad/s 2 (d) 12rad/s 2
9. A point on a flywheel of radius 0.5 m has
a uniform linear acceleration of 2 m/s 2 . Its
angular acceleration is:
(a) 2.5 rad/s 2 (b) 0.25 rad/s 2
(c) 1 rad/s 2 (d) 4 rad/s 2
Questions 10 to 13 refer to the following data.
A car accelerates uniformly from 10 m/s to
20 m/s over a distance of 150 m. The wheels of
the car each have a radius of 250 mm.
10. The time the car is accelerating is:
(a) 0.2 s (b) 15 s
(c) 10 s (d) 5 s
11. The initial angular velocity of each of the
wheels is:
(a) 20 rad/s (b) 40 rad/s
(c) 2.5 rad/s (d) 0.04 rad/s
(b) 67.5°
(d) 2.356°
Linear and angular motion 179
The angular acceleration of each of the
wheels is:
(a) 1 rad/s 2 (b) 0.25 rad/s 2
(c) 400 rad/s 2 (d) 4 rad/s 2
13. The linear acceleration of a point on each
of the wheels is:
(a) 1 m/s 2 (b) 4 m/s 2
(c) 3 m/s 2 (d) 100m/s 2
For fully worked solutions to each of the problems in Practice Exercises 73 to 78 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Three
Chapter 14
Why it is important to understand: Linear momentum and impulse
This chapter is of considerable importance in the study of the motion and collision of vehicles, ships,
and so on. The chapter commences with defining momentum, impulse, together with Newton’s laws of
motion. It then applies these laws to solving practical problems in the fields of ballistics, pile drivers,
etc. A study of linear and angular momentum is of importance when designing the motion and the crash
worthiness of cars, buses, etc.
At the end of this chapter you should be able to:
• define momentum and state its unit
• state Newton’s first law of motion
• calculate momentum given mass and velocity
• state Newton’s second law of motion
• define impulse and appreciate when impulsive forces occur
• state Newton’s third law of motion
• calculate impulse and impulsive force
• use the equation of motion v 2 = a 2 + las in calculations
14.1 Linear momentum
The momentum of a body is defined as the product of
its mass and its velocity, i.e.
momentum = mu
where m = mass (in kg) and u = velocity (in m/s). The
unit of momentum is kg m/s.
Since velocity is a vector quantity, momentum is
a vector quantity, i.e. it has both magnitude and
direction.
Newton’s first law of motion states:
a body continues in a state of rest or in a state of
uniform motion in a straight line unless acted
on by some external force
Hence the momentum of a body remains the same pro¬
vided no external forces act on it
The principle of conservation of momentum for a
closed system (i.e. one on which no external forces act)
may be stated as:
the total linear momentum of a system is a
constant
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
Linear momentum and impulse 181
The total momentum of a system before collision in a
given direction is equal to the total momentum of the
system after collision in the same direction. In Figure
14.1, masses m { and m 2 are travelling in the same direc¬
tion with velocity u { > u 2 . A collision will occur, and
applying the principle of conservation of momentum:
m-i
Figure 14.1
total momentum before impact = total momentum after
impact
i.e. + m 2 u 2 = + m 2 v 2
where Vj and v 2 are the velocities of m { and m 2 after
impact.
Determine the momentum of a pile
driver of mass 400 kg when it is moving downwards
with a speed of 12 m/s.
Momentum = mass x velocity = 400 kg x 12 m/s
= 4800 kg m/s downwards
A cricket ball of mass 150 g has a
momentum of 4.5 kg m/s. Determine the velocity
of the ball in km/h.
Momentum = mass x velocity, hence
, . momentum 4.5 kg m/s
velocity = -
mass 150 x 10 _3 kg
= 30 m/s
_ _ m _ _ s 1km
30 m/s = 30 — x 3600— x-
s h 1000m
- 30 x 3.6 km/h = 108 km/h
= velocity of cricket ball.
Determine the momentum of a rail¬
way wagon of mass 50 tonnes moving at a velocity
of 72 km/h.
Momentum = mass x velocity
Mass = 50 t = 50000 kg (since 1 t = 1000 kg)
and velocity = 72 km/h = 72 x ——— x 1000
h 3600s km
= —r m/s = 20 m/s.
3.6
Hence, momentum = 50000 kg x 20 m/s
= 1000000 kg m/s = 10 6 kg m/s
A wagon of mass 10 t is moving at
a speed of 6 m/s and collides with another wagon
of mass 15 t, which is stationary. After impact,
the wagons are coupled together. Determine the
common velocity of the wagons after impact.
Mass m { = 10 t = 10000 kg,
m 2 = 15000 kg and velocity u { = 6 m/s, u 2 = 0
Total momentum before impact = m ] u l + m 2 u 2
= (10000 x 6) + (15000 x 0) = 60000 kg m/s
Let the common velocity of the wagons after impact
be v m/s.
Since total momentum before impact = total momen¬
tum after impact:
60000 = m j v + m 2 v
= v(m^ + m 2 ) = v(25000)
Hence
60000
25000
= 2.4 m/s
i.e. the common velocity after impact is 2.4 m/s in
the direction in which the 10 t wagon is initially
travelling.
A body has a mass of 30 g and is
moving with a velocity of 20 m/s. It collides with
a second body which has a mass of 20 g and which
is moving with a velocity of 15 m/s. Assuming that
the bodies both have the same velocity after impact,
determine this common velocity, (a) when the initial
velocities have the same line of action and the same
sense, and (b) when the initial velocities have the
same line of action but are opposite in sense.
Mass m x = 30 g = 0.030 kg, m 2 — 20 g = 0.020 kg,
velocity u l =20 m/s and u 2 = 15 m/s.
(a) When the velocities have the same line of action
and the same sense, both u l and u 2 are considered
as positive values.
Total momentum before impact
= m { u { + m 2 u 2 = (0.030 x 20) + (0.020 x 15)
= 0.60 + 0.30 = 0.90 kg m/s.
Let the common velocity after impact be v m/s.
Total momentum before impact = total momen¬
tum after impact
Part Three
Part Three
182 Mechanical Engineering Principles
i.e. 0.90 = m x v + m 2 v = v(m x +m 2 )
0.90 = v(0.030 + 0.020)
from which, common velocity, v =
0.90
0.050
= 18 m/s
in the direction in which the bodies are initially
travelling.
When the velocities have the same line of action
but are opposite in sense, one is considered as
positive and the other negative (because velocity
is a vector quantity). Taking the direction of mass
m x as positive gives:
velocity u { = +20 m/s and u 2 = -15 m/s.
Total momentum before impact
= m x u x + m 2 u 2 = (0.030 x 20) + (0.020 x -15)
= 0.60-0.30 = + 0.30 kg m/s
and since it is positive this indicates a momentum
in the same direction as that of mass m j
If the common velocity after impact is v m/s then
0.30 = v(m { + m 2 ) = v(0.050)
from which, common velocity, v = = 6 m/s
J 0.050
in the direction that the 30 g mass is initially
travelling.
A ball of mass 50 g is moving with
a velocity of 4 m/s when it strikes a stationary ball
of mass 25 g. The velocity of the 50 g ball after
impact is 2.5 m/s in the same direction as before
impact. Determine the velocity of the 25 g ball
after impact.
Mass m l = 50 g = 0.050 kg, m 2 = 25 g = 0.025 kg.
Initial velocity u { = 4 m/s, u 2 = 0;
final velocity v x = 2.5 m/s, v 2 is unknown.
Total momentum before impact
= m { u l + rn 2 u 2
= (0.050 x 4) + (0.025 x 0) = 0.20 kg m/s
Total momentum after impact
= AWjVj + m 2 V 2
= (0.050 x 2.5) + (0.025 v 2 ) = 0.125 + 0.025 v 2
Total momentum before impact = total momentum
after impact, hence
0.20 = 0.125 + 0.025v 2
from which, velocity of 25 g ball after impact,
0.20-0.125
V 2 = 0.025 =3 m/S
Three masses, P, Q and R lie in a
straight line. P has a mass of 5 kg and is moving
towards Q at 8 m/s. Q has a mass of 7 kg and a
velocity of 4 m/s, and is moving towards R. Mass
R is stationary. P collides with Q, and P and Q then
collide with R. Determine the mass of R assuming
all three masses have a common velocity of 2 m/s
after the collision of P and Q with R.
Mass m p = 5 kg, = 7 kg,
velocity u p = 8 m/s and Uq = 4 m/s.
Total momentum before P collides with Q
= m p u p + MqUq
= (5 x 8) + (7 x 4) = 68 kg m/s.
Let P and Q have a common velocity of v x m/s after
impact.
Total momentum after P and Q collide
= m p v x + mgV x
= Vj (m p + m Q ) = 12vj
Total momentum before impact = total momentum
after impact, i.e. 68 = 12v 1? from which, common
68 2
velocity of P and Q,v x = — = 5^ m/s.
Total momentum after P and Q collide with R =
(ntp+Q x 2) + (m R x 2) (since the common velocity after
impact = 2 m/s) = (12 x 2) + (2 m R )
Total momentum before P and Q collide with R = total
momentum after P and Q collide with R , i.e.
2
(jn p + q x 5 —) = (12 x 2) + 2 m R
2
i.e. 12 x 5 — = 24 + 2 m R
68 — 24 = 2 m R
from which, mass of R , m R =
44
T
= 22 kg.
Now try the following Practice Exercise
Practice Exercise 79 Further problems on
linear momentum
(Where necessary, take g as 9.81 m/s 2 )
1. Determine the momentum in a mass of
50 kg having a velocity of 5 m/s.
[250 kg m/s]
Linear momentum and impulse 183
2. A milling machine and its component have
a combined mass of 400 kg. Determine the
momentum of the table and component
when the feed rate is 360 mm/min.
[2.4 kg m/s]
3. The momentum of a body is 160 kg m/s
when the velocity is 2.5 m/s. Determine the
mass of the body. [64 kg]
4. Calculate the momentum of a car of mass
750 kg moving at a constant velocity of
108 km/h. [22500 kg m/s]
5. A football of mass 200 g has a momentum
of 5 kg m/s. What is the velocity of the ball
in km/h. [90 km/h]
6 . A wagon of mass 8 t is moving at a speed
of 5 m/s and collides with another wagon
of mass 12 t, which is stationary. After
impact, the wagons are coupled together.
Determine the common velocity of the
wagons after impact. [2 m/s]
7. A car of mass 800 kg was stationary when
hit head-on by a lorry of mass 2000 kg
travelling at 15 m/s. Assuming no brakes
are applied and the car and lorry move as
one, determine the speed of the wreckage
immediately after collision. [10.71 m/s]
8 . A body has a mass of 25 g and is moving
with a velocity of 30 m/s. It collides with a
second body which has a mass of 15 g and
which is moving with a velocity of 20 m/s.
Assuming that the bodies both have the
same speed after impact, determine their
common velocity (a) when the speeds have
the same line of action and the same sense,
and (b) when the speeds have the same line
of action but are opposite in sense.
[(a) 26.25 m/s (b) 11.25 m/s]
9. A ball of mass 40 g is moving with a veloc¬
ity of 5 m/s when it strikes a stationary ball
of mass 30 g. The velocity of the 40 g ball
after impact is 4 m/s in the same direction
as before impact. Determine the velocity of
the 30 g ball after impact. [1.33 m/s]
10. Three masses, X , 7 and Z, lie in a straight
line. X has a mass of 15 kg and is mov¬
ing towards Y at 20 m/s. Y has a mass of
10 kg and a velocity of 5 m/s and is moving
towards Z. Mass Z is stationary. X collides
with 7, and X and 7 then collide with Z.
Determine the mass of Z assuming all three
masses have a common velocity of 4 m/s
after the collision of X and 7 with Z.
[62.5 kg]
14.2 Impulse and impulsive forces
Newton’s second law of motion states:
the rate of change of momentum is directly
proportional to the applied force producing the
change , and takes place in the direction of this
force
In the SI system, the units are such that:
the applied force = rate of change of momentum
change of momentum
time taken
When a force is suddenly applied to a body due to
either a collision with another body or being hit by an
object such as a hammer, the time taken in equation
(14.1) is very small and difficult to measure. In such
cases, the total effect of the force is measured by the
change of momentum it produces.
Forces that act for very short periods of time are
called impulsive forces. The product of the impul¬
sive force and the time during which it acts is called
the impulse of the force and is equal to the change of
momentum produced by the impulsive force, i.e.
impulse = applied force x time
= change in linear momentum
Examples where impulsive forces occur include when
a gun recoils and when a free-falling mass hits the
ground. Solving problems associated with such occur¬
rences often requires the use of the equation of motion:
v 2 = u 2 + las , from Chapter 13.
When a pile is being hammered into the ground,
the ground resists the movement of the pile and this
resistance is called a resistive force.
Newton’s third law of motion may be stated as:
for every action there is an equal and opposite
reaction
The force applied to the pile is the resistive force; the
pile exerts an equal and opposite force on the ground.
Part Three
Part Three
184 Mechanical Engineering Principles
In practice, when impulsive forces occur, energy is not
entirely conserved and some energy is changed into
heat, noise, and so on.
The average force exerted on the
work-piece of a press-tool operation is 150 kN,
and the tool is in contact with the work-piece
for 50 ms. Determine the change in momentum.
From above, change of linear momentum = applied
force x time (= impulse).
Hence, change in momentum of work-piece
= 150 x 10 3 N x 50 x 10 _3 s
= 7500 kg m/s (since 1 N = 1 kg m/s 2 ).
The hammer of a pile-driver of
mass 1 t falls a distance of 1.5 m on to a pile. The
blow takes place in 25 ms and the hammer does
not rebound. Determine the average applied force
exerted on the pile by the hammer.
Initial velocity, u = 0,
acceleration due to gravity, g = 9.81 m/s 2
and distance, 5=1.5m
Using the equation of motion: v 2 = u 2 + 2 gs
gives: v 2 = 0 2 + 2(9.81)(1.5)
from which, impact velocity, v = - N /(2)(9.81)(1.5)
= 5.425 m/s
A force of 15 N acts on a body of mass
4 kg for 0.2 s. Determine the change in velocity.
Impulse = applied force x time
= change in linear momentum
i.e. 15 N x 0.2 s = mass x change in velocity
= 4 kg x change in velocity
from which, change in velocity
_ 15N x 0.2s
4 kg
= 0.75 m/s (since 1 N = 1 kg m/s 2 ).
A mass of 8 kg is dropped vertically
on to a fixed horizontal plane and has an impact
velocity of 10 m/s. The mass rebounds with a
velocity of 6 m/s. If the mass-plane contact time
is 40 ms, calculate (a) the impulse, and (b) the
average value of the impulsive force on the plane.
(a) Impulse = change in momentum = m(u { - v { )
where u { = impact velocity = 10 m/s and
Vj = rebound velocity = -6 m/s
(vj is negative since it acts in the opposite direc¬
tion to Wj, and velocity is a vector quantity)
Thus, impulse = m(u l - Vj) = 8 kg (10 —6) m/s
= 8 x 16 = 128 kg m/s.
impulse 128 kg m/s
(b) Impulsive force = —:-=-—
time 40 x 10 _3 s
= 3200 N or 3.2 kN
Neglecting the small distance moved by the pile and
hammer after impact,
momentum lost by hammer = the change of momentum
= mv= 1000 kg x 5.425 m/s
Rate of change of momentum
change of momentum
change of time
1000 x 5.425
25 x 10 -3
- 217000N
Since the impulsive force is the rate of change of
momentum, the average force exerted on the pile is
217 kN.
A mass of 40 g having a velocity of
15 m/s collides with a rigid surface and rebounds
with a velocity of 5 m/s. The duration of the impact
is 0.20 ms. Determine (a) the impulse, and (b) the
impulsive force at the surface.
Mass m = 40 g = 0.040 kg,
initial velocity, u= 15 m/s
and final velocity, v = -5 m/s (negative since the
rebound is in the opposite
direction to velocity u)
(a) Momentum before impact = mu = 0.040 x 15
= 0.6 kg m/s
Momentum after impact = mv = 0.040 x - 5
= - 0.2 kg m/s
Impulse = change of momentum
= 0.6 - (-0.2) = 0.8 kg m/s
Linear momentum and impulse 185
(b) Impulsive force
__ change of momentum
change of time
= ° 8kgm,S — 4000 N or 4 kN
0.20 x 10 _3 s
A gun of mass 1.5 t fires a shell of
mass 15 kg horizontally with a velocity of 500 m/s.
Determine (a) the initial velocity of recoil, and
(b) the uniform force necessary to stop the recoil
of the gun in 200 mm.
Mass of gun, m = 1.5 t = 1500 kg,
o
mass of shell, m s = 15 kg,
and initial velocity of shell, u s = 500 m/s.
(a) Momentum of shell = mu = 15 x 500
= 7500 kg m/s.
Momentum of gun = m v = 1500v
o
where v = initial velocity of recoil of the gun.
By the principle of conservation of momentum,
initial momentum = final momentum, i.e.
0 = 7500 + 1500v, from which,
velocity v = - = - 5 m/s (the negative sign
indicating recoil velocity)
i.e. the initial velocity of recoil = 5 m/s.
(b) The retardation of the recoil, a, may be deter¬
mined using v 2 = u 2 + las , where v, the final
velocity, is zero, u, the initial velocity, is 5 m/s
and s, the distance, is 200 mm, i.e. 0.2 m.
Rearranging v * 1 2 = u 2 + las for a gives:
v 2 -u 2 _ 0 2 -5 2
a= Is ~ 2(0.2)
= — = -62.5 m/s 2
0.4
Force necessary to stop recoil in 200 mm
= mass x acceleration
= 1500 kg x 62.5 m/s 2
= 93750 N or 93.75 kN
A vertical pile of mass 100 kg is
driven 200 mm into the ground by the blow of a 1 t
hammer which falls through 750 mm. Determine
(a) the velocity of the hammer just before impact
(b) the velocity immediately after impact (assuming
the hammer does not bounce), and (c) the resistive
force of the ground assuming it to be uniform.
For the hammer, v 2 = u 2 + Igs , where v = final
velocity, u = initial velocity = 0, g = 9.81 m/s 2
and s = distance = 750 mm = 0.75 m
Hence, v 2 = 0 2 + 2(9.81)(0.75), from which,
velocity of hammer, just before impact,
v = ^2(9.81X0.75) =3.84 m/s
Momentum of hammer just before impact
= mass x velocity
= 1000 kg x 3.84 m/s = 3840 kg m/s
Momentum of hammer and pile after impact =
momentum of hammer before impact.
Hence, 3840 kg m/s = (mass of hammer and pile)
x (velocity immediately after impact)
i.e. 3840 = (1000 + 100)(v), from which,
velocity immediately after impact,
3840
1100
= 3.49 m/s
Resistive force of ground = mass x acceleration.
The acceleration is determined using v 2 = u 2 + las
where v = final velocity = 0, u = initial velocity =
3.49 m/s and s = distance driven in ground =
200 mm = 0.2 m.
Hence, 0 2 = (3.49) 2 + 2(a)(0.2), from which,
acceleration, a = = -30.45 m/s 2 (the
2 ( 0 . 2 )
minus sign indicates retardation, because accel¬
eration is a vector quantity).
Thus, resistive force of ground
= mass x acceleration
= 1100 kg x 30.45 m/s 2 = 33.5 kN
Now try the following Practice Exercises
Practice Exercise 80 Further problems
on impulse and
impulsive forces
(Where necessary, take g as 9.81 m/s 2 )
1. The sliding member of a machine tool has
a mass of 200 kg. Determine the change
in momentum when the sliding speed is
increased from 10 mm/s to 50 mm/s.
[8 kg m/s]
2. A force of 48 N acts on a body of mass 8 kg
for 0.25 s. Determine the change in velocity.
[1.5 m/s]
Part Three
Part Three
186 Mechanical Engineering Principles
3. The speed of a car of mass 800 kg is
increased from 54 km/h to 63 km/h in 2 s.
Determine the average force in the direc¬
tion of motion necessary to produce the
change in speed. [1 kN]
4. A 10 kg mass is dropped vertically on to
a fixed horizontal plane and has an impact
velocity of 15 m/s. The mass rebounds with
a velocity of 5 m/s. If the contact time of
mass and plane is 0.025 s, calculate (a) the
impulse, and (b) the average value of the
impulsive force on the plane.
[(a) 200 kg m/s (b) 8 kN]
5. The hammer of a pile driver of mass 1.2 t
falls 1.4 m on to a pile. The blow takes
place in 20 ms and the hammer does not
rebound. Determine the average applied
force exerted on the pile by the hammer.
[314.5 kN]
6 . A tennis ball of mass 60 g is struck from
rest with a racket. The contact time of
ball on racket is 10 ms and the ball leaves
the racket with a velocity of 25 m/s.
Calculate (a) the impulse, and (b) the aver¬
age force exerted by a racket on the ball.
[(a) 1.5 kg m/s (b) 150 N]
7. In a press-tool operation, the tool is in
contact with the work piece for 40 ms. If
the average force exerted on the work
piece is 90 kN, determine the change in
momentum. [3600 kg m/s]
8 . A gun of mass 1.21 fires a shell of mass 12 kg
with a velocity of 400 m/s. Determine
(a) the initial velocity of recoil, and (b) the
uniform force necessary to stop the recoil
of the gun in 150 mm.
[(a) 4 m/s (b) 64 kN]
9. In making a steel stamping, a mass of
100 kg falls on to the steel through a dis¬
tance of 1.5 m and is brought to rest af¬
ter moving through a further distance of
15 mm. Determine the magnitude of the re¬
sisting force, assuming a uniform resistive
force is exerted by the steel. [98.1 kN]
10. A vertical pile of mass 150 kg is driven
120 mm into the ground by the blow of a 1.11
hammer, which falls through 800 mm.
Assuming the hammer and pile remain in
contact, determine (a) the velocity of the
hammer just before impact (b) the velocity
immediately after impact, and (c) the resis¬
tive force of the ground, assuming it to be
uniform.
[(a) 3.96 m/s (b) 3.48 m/s (c) 63.08 kN]
Practice Exercise 81 Short-answer
questions on linear
momentum and
impulse
1. Define momentum.
2. State Newton’s first law of motion.
3. State the principle of the conservation of
momentum.
4. State Newton’s second law of motion.
5. Define impulse.
6 . What is meant by an impulsive force?
7. State Newton’s third law of motion.
Practice Exercise 82 Multiple-choice
questions on linear
momentum and
impulse
(Answers on page 335)
1. A mass of 100 g has a momentum of
100 kg m/s. The velocity of the mass is:
(a) 10 m/s (b) 10 2 m/s
(c) 10 -3 m/s (d) 10 3 m/s
2. A rifle bullet has a mass of 50 g. The
momentum when the muzzle velocity is
108 km/h is:
(a) 54 kg m/s (b) 1.5 kg m/s
(c) 15000 kg m/s (d) 21.6 kg m/s
A body P of mass 10 kg has a velocity of
5 m/s and the same line of action as a body
Q of mass 2 kg and having a velocity of
25 m/s. The bodies collide, and their veloc¬
ities are the same after impact. In questions
Linear momentum and impulse 187
3 to 6, select the correct answer from the
following:
(a) 25/3 m/s (b) 360 kg m/s
(c) 0 (d) 30 m/s
(e) 160 kg m/s (f) 100 kg m/s
(g) 20 m/s
3. Determine the total momentum of the
system before impact when P and Q have
the same sense.
4. Determine the total momentum of the
system before impact when P and Q have
the opposite sense.
5. Determine the velocity of P and Q after im¬
pact if their sense is the same before impact.
6 . Determine the velocity of P and Q after
impact if their sense is opposite before
impact.
7. A force of 100 N acts on a body of mass
10 kg for 0.1 s. The change in velocity of
the body is:
(a) 1 m/s (b) 100 m/s
(c) 0.1 m/s (d) 0.01 m/s.
A vertical pile of mass 200 kg is driven
100 mm into the ground by the blow of a
1 t hammer which falls through 1.25 m.
In questions 8 to 12, take g as 10 m/s 2 and
select the correct answer from the following:
(a) 25 m/s
(c) 5 kg m/s
(e) 625/6 kN
(g) 5 m/s
(b) 25/6 m/s
(d) 0
(f) 5000 kg m/s
(h) 12 kN
8 . Calculate the velocity of the hammer im¬
mediately before impact.
9. Calculate the momentum of the hammer
just before impact.
10. Calculate the momentum of the hammer
and pile immediately after impact assum¬
ing they have the same velocity.
11. Calculate the velocity of the hammer and
pile immediately after impact assuming
they have the same velocity.
12. Calculate the resistive force of the ground,
assuming it to be uniform.
For fully worked solutions to each of the problems in Practice Exercises 79 to 82 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Three
Chapter 15
Why it is important to understand: Force, mass and acceleration
When an object is pushed or pulled, a force is applied to the object. The effects of pushing or pulling an
object are to cause changes in the motion and shape of the object. If a change occurs in the motion
of the object then the object accelerates. Thus, acceleration results from a force being applied to an
object. If a force is applied to an object and it does not move, then the object changes shape. Usually
the change in shape is so small that it cannot be detected by just watching the object. However, when
very sensitive measuring instruments are used, very small changes in dimensions can be detected.
A force of attraction exists between all objects. If a person is taken as one object and the Earth as a
second object, a force of attraction exists between the person and the Earth. This force is called the
gravitational force and is the force that gives a person a certain weight when standing on the Earth’s
surface. It is also this force that gives freely falling objects a constant acceleration in the absence of
other forces. This chapter defines force and acceleration, states Newton’s three laws of motion and
defines moment of inertia, all demonstrated via practical everyday situations. The importance of this
chapter is in understanding the dynamic behaviour of moving motor vehicles, ships, etc.
At the end of this chapter you should be able to:
• define force and state its unit
• appreciate ‘gravitational force’
• state Newton’s three laws of motion
• perform calculations involving force F = ma
• define ‘centripetal acceleration’
• perform calculations involving centripetal force =
• define ‘mass moment of inertia’
15.1 Introduction
As stated above, when an object is pushed or pulled, a
force is applied to the object. This force is measured
in newtons (N). The effects of pushing or pulling an
object are:
(i) to cause a change in the motion of the object, and
(ii) to cause a change in the shape of the object.
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
Force, mass and acceleration 189
If a change occurs in the motion of the object, that is,
its velocity changes from u to v, then the object accel¬
erates. Thus, it follows that acceleration results from a
force being applied to an object. If a force is applied to
an object and it does not move, then the object changes
shape, that is, deformation of the object takes place.
Usually the change in shape is so small that it cannot be
detected by just watching the object. However, when
very sensitive measuring instruments are used, very
small changes in dimensions can be detected.
A force of attraction exists between all objects. The
factors governing the size of this force F are the masses
of the objects and the distances between their centres:
Fa
m\ni2
Thus, if a person is taken as one object and the
Earth as a second object, a force of attraction exists
between the person and the Earth. This force is called
the gravitational force, (first presented by Sir Isaac
Newton*), and is the force that gives a person a certain
weight when standing on the Earth’s surface. It is also
* sir Isaac Newton (25 December 1642-20 March 1727) was an
English polymath. His single greatest work, the ‘Philosophiae
Naturalis Principia Mathematica’ (‘Mathematical Principles of
Natural Philosophy’), showed how a universal force, gravity,
applied to all objects in all parts of the universe. To find out
more go to www.routledge.com/cw/bird
this force that gives freely falling objects a constant
acceleration in the absence of other forces.
15.2 Newton's laws of motion
To make a stationary object move or to change the direc¬
tion in which the object is moving requires a force to be
applied externally to the object. This concept is known
as Newton’s first law of motion and may be stated as:
An object remains in a state of rest , or contin¬
ues in a state of uniform motion in a straight
line , unless it is acted on by an externally ap¬
plied force
Since a force is necessary to produce a change of
motion, an object must have some resistance to a change
in its motion. The force necessary to give a stationary
pram a given acceleration is far less than the force
necessary to give a stationary car the same acceleration
on the same surface. The resistance to a change in motion
is called the inertia of an object and the amount of
inertia depends on the mass of the object. Since a car
has a much larger mass than a pram, the inertia of a car
is much larger than that of a pram.
Newton’s second law of motion may be stated as:
The acceleration of an object acted upon by an
external force is proportional to the force and is
in the same direction as the force
Thus, force a acceleration, or force = a constant x
acceleration, this constant of proportionality being the
mass of the object, i.e.
force = mass x acceleration
The unit of force is the newton (N) and is defined
in terms of mass and acceleration. One newton is
the force required to give a mass of 1 kilogram an
acceleration of 1 metre per second squared. Thus
F = ma
where F is the force in newtons (N), m is the mass in
kilograms (kg) and a is the acceleration in metres per
second squared (m/s 2 ), i.e. 1 N
lkgm
It follows that 1 m/s 2 = 1 N/kg. Hence a gravitational
acceleration of 9.8 m/s 2 is the same as a gravitational
field of 9.8 N/kg.
Newton’s third law of motion may be stated as:
For every force , there is an equal and opposite
reacting force
Part Three
Part Three
190 Mechanical Engineering Principles
Thus, an object on, say, a table, exerts a downward
force on the table and the table exerts an equal upward
force on the object, known as a reaction force or just
a reaction.
Calculate the force needed to
accelerate a boat of mass 20 tonne uniformly
from rest to a speed of 21.6 km/h in 10 minutes.
The mass of the boat, m , is 20 t, that is 20000 kg.
The law of motion, v = u + at can be used to determine
the acceleration a.
The initial velocity, u, is zero, the final velocity,
v
., km
= 21.6 km/h = 21 . 6 ——x
lh
x
1000 m
h 3600s 1 km
21.6
~F6
= 6 m/s,
and the time, t = 10 min = 600 s.
Thus, v = u + at i.e. 6 = 0 + <2 x 600
6 9
from which, a = = 0.01 m/s z
600
From Newton’s second law, F= ma
i.e. force = 20000 x 0.01 N = 200 N
The moving head of a machine tool
requires a force of 1.2 N to bring it to rest in 0.8 s
from a cutting speed of 30 m/min. Find the mass of
the moving head.
F
From Newton’s second law, F= ma, thus m = —, where
force is given as 1.2 N. The law of motion
v = w + at can be used to find acceleration a ,
30
where v = 0, u = 30 m/min = — m/s = 0.5 m/s,
60
and t = 0.8 s.
Thus, 0 = 0.5+ < 2 x 0.8
from which, a = — = -0.625 m/s 2 or a retardation
0.8
of 0.625 m/s 2 .
Thus the mass, m =
F
a
1.2
0.625
= 1.92 kg
A lorry of mass 1350 kg accelerates
uniformly from 9 km/h to reach a velocity of
45 km/h in 18 s. Determine (a) the acceleration of
the lorry (b) the uniform force needed to accelerate
the lorry.
(a) The law of motion v = u + at can be used to deter¬
mine the acceleration, where final velocity
45
. _km
v = 45— x
lh 1000 m
x
h 3600s 1km
3.6
m/s, initial
velocity u = — m/s and time t— 18 s.
3.6
Thus
45
3.6 3.6
1
+ <2X18
from which, <2 =
45
9 ^
1
36 >
[ 3.6
3 . 6 ,
18
[ 3.6 J
18
= — = — m/s 2 or 0.556 m/s 2
18 9
(b) From Newton’s second law of motion,
force, F = ma = 1350 x — = 750 N
9
Find the weight of an object of mass
1 .6 kg at a point on the Earth’s surface where the
gravitational field is 9.81 N/kg (or 9.81 m/s 2 ).
The weight of an object is the force acting vertically
downwards due to the force of gravity acting on the
object.
Thus: weight = force acting vertically downwards
= mass x gravitational field
= 1.6x9.81 = 15.696 N
A bucket of cement of mass 40 kg
is tied to the end of a rope connected to a hoist.
Calculate the tension in the rope when the bucket
is suspended but stationary. Take the gravitational
field, g , as 9.81 N/kg (or 9.81 m/s 2 ).
The tension in the rope is the same as the force acting
in the rope. The force acting vertically downwards due
to the weight of the bucket must be equal to the force
acting upwards in the rope, i.e. the tension.
Weight of bucket of cement, F = mg = 40 x 9.81
= 392.4 N
Thus, the tension in the rope = 392.4 N
The bucket of cement in Problem 5
is now hoisted vertically upwards with a uniform
acceleration of 0.4 m/s 2 . Calculate the tension in the
rope during the period of acceleration.
Force, mass and acceleration 191
With reference to Figure 15.1, the forces acting on the
bucket are:
(i) a tension (or force) of T acting in the rope
(ii) a force of mg acting vertically downwards, i.e.
the weight of the bucket and cement
A 7 "
By comparing this result with that of Problem 5,
it can be seen that there is a decrease in the ten¬
sion in the rope when an object is accelerating
downwards.
Now try the following Practice Exercise
A
Acceleration
Weight,
mg
Force due
to acceleration
F-ma
Figure 15.1
The resultant force F= T- mg
Hence, ma = T -mg
i.e. 40 x 0.4 = T- 40 x 9.81
from which, tension, T = 408.4 N
By comparing this result with that of Problem 5, it
can be seen that there is an increase in the tension
in the rope when an object is accelerating upwards.
The bucket of cement in Problem 5 is
now lowered vertically downwards with a uniform
acceleration of 1.4 m/s 2 . Calculate the tension in the
rope during the period of acceleration.
With reference to Figure 15.2, the forces acting on the
bucket are:
(i) a tension (or force) of T acting vertically upwards
T u F=ma
Acceleration
V
Weight,
mg
Figure 15.2
(ii) a force of mg acting vertically downwards, i.e.
the weight of the bucket and cement
The resultant force, F = mg - T
Hence, ma = mg - T
from which, tension, T—m(g —a)
= 40(9.81 - 1.4)
Practice Exercise 83 Further problems
on Newton's laws of
motion
(Take g as 9.81 m/s 2 , and express answers to
three significant figure accuracy)
1. A car initially at rest, accelerates uniformly
to a speed of 55 km/h in 14 s. Determine the
accelerating force required if the mass of the
car is 800 kg. [873 N]
2. The brakes are applied on the car in question
1 when travelling at 55 km/h and it comes to
rest uniformly in a distance of 50 m. Calcu¬
late the braking force and the time for the car
to come to rest. [1.87 kN, 6.55 s]
3. The tension in a rope lifting a crate ver¬
tically upwards is 2.8 kN. Determine its
acceleration if the mass of the crate is 270 kg.
[0.560 m/s 2 ]
4. A ship is travelling at 18 km/h when it stops
its engines. It drifts for a distance of 0.6 km
and its speed is then 14 km/h. Determine the
value of the forces opposing the motion of
the ship, assuming the reduction in speed is
uniform and the mass of the ship is 2000 t.
[16.5 kN]
5. A cage having a mass of 2 t is being lowered
down a mineshaft. It moves from rest with
an acceleration of 4 m/s 2 , until it is travelling
at 15 m/s. It then travels at constant speed
for 700 m and finally comes to rest in 6 s.
Calculate the tension in the cable supporting
the cage during
(a) the initial period of acceleration
(b) the period of constant speed travel
(c) the final retardation period.
[(a) 11.6 kN (b) 19.6 kN (c) 24.6 kN]
6. A miner having a mass of 80 kg is standing
in the cage of Problem 5. Determine the
reaction force between the man and the floor
= 336.4 N
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192 Mechanical Engineering Principles
of the cage during (a) the initial period of
acceleration (b) the period of constant speed
travel, and (c) the final retardation period,
[(a) 464.8 N (b) 784.8 N (c) 984.8 N]
7. During an experiment, masses of 4 kg and
5 kg are attached to a thread and the thread
is passed over a pulley so that both masses
hang vertically downwards and are at the
same height. When the system is released,
find (a) the tension in the thread, and (b)
the acceleration of the system, assuming no
losses in the system.
[(a) 43.6 N (b) 1.09 m/s 2 ]
15.3 Centripetal acceleration
When an object moves in a circular path at constant
speed, its direction of motion is continually changing
and hence its velocity (which depends on both magni¬
tude and direction) is also continually changing. Since
acceleration is the (change in velocity)/(time taken) the
object has an acceleration.
Let the object be moving with a constant angular
velocity of co and a tangential velocity of magnitude
v and let the change of velocity for a small change
of angle of 0 (= cot) be V (see Figure 15.3(a)). Then,
v 2 -vj = V
Figure 15.3
The vector diagram is shown in Figure 15.3(b) and
since the magnitudes of Vj and v 2 are the same, i.e. v,
the vector diagram is also an isosceles triangle.
Bisecting the angle between v 2 and gives:
. 0 V/2 V
sin — = -= —
2 v 2 2v
i.e.
V — 2v sin —
2
(15.1)
Since
0 = cot , then t = —
Q_
co
(15.2)
Dividing (15.1) by (15.2) gives:
, • o
V 2v Sln 2
t
0
CO
. e
vmsin —
_2
0
2
. 0
sin —
2
For small angles, ~^ is very nearly equal to unity
Therefore,
or,
V
— = vco
t
V _ change of velocity
t change of time
= acceleration, a = vco
v
But, co = v/r, thus vco = v x —
r
v
= acceleration
v
That is, the acceleration a is — and is towards the
r
centre of the circle of motion (along V). It is called the
centripetal acceleration. If the mass of the rotating
object is m, then by Newton’s second law, the centrip-
,2
etal force is
mv
, and its direction is towards the cen¬
tre of the circle of motion.
A vehicle of mass 750 kg travels round
a bend of radius 150 m, at 50.4 km/h. Determine the
centripetal force acting on the vehicle.
The centripetal force is given by
r
is towards the centre of the circle.
and its direction
and
m = 750 kg,
v = 50.4 km/h =
r= 150 m
50.4
m/s = 14 m/s
Thus, centripetal force
750x14 2
150
980 N
An object is suspended by a thread
250 mm long and both object and thread move in
a horizontal circle with a constant angular velocity
of 2.0 rad/s. If the tension in the thread is 12.5 N,
determine the mass of the object.
Force, mass and acceleration 193
Centripetal force (i.e. tension in thread) =
mv * 1 2
r
= 12.5 N
The angular velocity, co = 2.0 rad/s
and radius, r = 250 mm = 0.25 m.
Since linear velocity v = cor , v = 2.0 x 0.25
= 0.5 m/s,
mv
and since F = -, then m =
Fr
v
i.e. mass of object, m
12.5x0.25
0.5 2
12.5 kg
An aircraft is turning at constant alti¬
tude, the turn following the arc of a circle of radius
1.5 km. If the maximum allowable acceleration of
the aircraft is 2.5 g, determine the maximum speed
of the turn in km/h. Take g as 9.8 m/s 2 .
The acceleration of an object turning in a circle is
Thus, to determine the maximum speed of turn
Hence, speed of turn, v = yj 2.5gr = V2.5 x 9.8 x 1500
= ^36750 = 191.7 m/s
= 191.7x3.6 km/h
= 690 km/h
Now try the following Practice Exercise
Practice Exercise 84 Further problems
on centripetal
acceleration
1. Calculate the centripetal force acting on
a vehicle of mass 1 tonne when travelling
round a bend of radius 125 m at 40 km/h. If
this force should not exceed 750 N, deter¬
mine the reduction in speed of the vehicle to
meet this requirement. [988 N, 34.86 km/h]
2. A speed-boat negotiates an S-bend consisting
of two circular arcs of radii 100 m and
150 m. If the speed of the boat is constant at
34 km/h, determine the change in acceleration
when leaving one arc and entering the other.
[0.3 m/s 2 ]
3. An object is suspended by a thread 400 mm
long and both object and thread move in a
horizontal circle with a constant angular
velocity of 3.0 rad/s. If the tension in the
thread is 36 N, determine the mass of the
object. [10 kg]
15.4 Rotation of a rigid body
about a fixed axis
A rigid body is said to be a body that does not change
its shape or size during motion. Thus, any two particles
on a rigid body will remain the same distance apart
during motion.
Consider the rigidity of Figure 15.4, which is rotating
about the fixed axis O.
Figure 15.4
In Figure 15.4,
a = the constant angular acceleration
m = the mass of a particle
r = the radius of rotation of m
a t = the tangential acceleration of m
F t = the elemental force on the particle
Now, force F = ma
or F t = m a (
= m ( ar )
Multiplying both sides of the above equation by r,
gives:
F ( r= mar 2
Since a is a constant
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194 Mechanical Engineering Principles
But T= F { r
and mr 2 = I
Hence, T=I o a (15.3)
where T = the total turning moment exerted on the
rigid body = ^ F t r
and I Q = the mass moment of inertia (or second
moment) about O (in kg m 2 ).
Equation (15.3) can be seen to be the rotational equiva¬
lent of F = ma (Newton’s second law of motion).
Determine the angular acceleration
that occurs when a circular disc of mass moment
of inertia of 0.5 kg m 2 is subjected to a torque of
6 N m. Neglect friction and other losses.
From equation (15.3), torque T= la,
from which, angular acceleration, a = — = — .
I 0.5 kg m 2
= 12 rad/s 2
15.5 Moment of inertia (/)
The moment of inertia is required for analysing
problems involving the rotation of rigid bodies. It is
defined as:
/ = mk 2 = mass moment of inertia (kg/m 2 )
or moment of inertia,
where m = the mass of the rigid body
k = its radius of gyration about the point of
rotation (see Chapter 9).
In general, / = mr2 where the definitions of
Figure 15.4 apply.
Some typical values of mass and the radius of
gyration are given in Table 15.1, where
A = cross-sectional area
L = length
t = disc thickness
R = radius of the solid disc
R j = internal radius
R 2 = external radius
p = density
Table 15.1
Component
mass
k 2
Rod, about
pAL
L 2
mid-point
12
Rod, about
pAL
L 2
an end
3
Flat disc
pnR 2 t
R 2
2
(R 2 + R 2 )
Annulus
p7i{R 2 - R 2 )t
2
Parallel axis theorem
This is of similar form to the parallel axis theorem of
Chapter 9, where
I xx = I G + mh2
/ = the mass moment of inertia about the xx axis,
which is parallel to an axis passing through
the centre of gravity of the rigid body, namely
at G
Iq = the mass moment of inertia of the rigid body
about an axis passing through G and parallel to
the xx axis
h = the perpendicular distance between the above
two parallel axes.
Determine the mass moment
of inertia about its centroid for a solid uniform
thickness disc. For the disc, its radius is 0.2 m, its
thickness is 0.05 m, and its density is 7860 kg/m 3 .
From Table 15.1, for a disc,
mass, m = pnR 2 t
= 7860 x jz x (0.2 m) 2 x 0.05 m
ITT
= 49.386 kg
Mass moment of inertia about its centroid,
49.386 x 0.2 2 , ,
-kgm
2 s
= 0.988 kg m 2
Force, mass and acceleration 195
Now try the following Practice Exercises
Practice Exercise 85 Further problems on
rotation and moment
of inertia
1. Calculate the mass moment of inertia of a
thin rod, of length 0.5 m and mass 0.2 kg,
about its centroid. [0.004167 kg m 2 ]
2. Calculate the mass moment of inertia of the
thin rod of Problem 1, about an end.
[0.01667 kg m 2 ]
3. Calculate the mass moment of inertia of a
solid disc of uniform thickness about its
centroid. The diameter of the disc is 0.3 m
and its thickness is 0.08 m. The density of its
material of construction is 7860 kg/m 3 .
[0.50 kg m 2 ]
4. If a hole of diameter 0.2 m is drilled through
the centre of the disc of Problem 3, what
will be its mass moment of inertia about its
centroid? [0.401 kg m 2 ]
Practice Exercise 86 Short-answer
questions on force,
mass and acceleration
1. Force is measured in.
2. The two effects of pushing or pulling an
object are.or.
3. A gravitational force gives free-falling ob¬
jects a.in the absence of all other forces.
4. State Newton’s first law of motion.
5. Describe what is meant by the inertia of an
object.
6. State Newton’s second law of motion.
7. Define the newton.
8. State Newton’s third law of motion.
9. Explain why an object moving round a
circle at a constant angular velocity has an
acceleration.
10. Define centripetal acceleration in symbols.
11. Define centripetal force in symbols.
12. Define mass moment of inertia.
13. A rigid body has a constant angular accel¬
eration a when subjected to a torque T. The
mass moment of inertia, I = .
Practice Exercise 87 Multiple-choice
questions on force,
mass and acceleration
(Answers on page 336)
1. The unit of force is the:
(a) watt (b) kelvin
(c) newton (d) joule
2. If a = acceleration and F = force, then mass
m is given by:
F
(a) m = a - F (b) m = —
a
(c) m= F-a (d) m = —
F
3. The weight of an object of mass 2 kg at a
point on the Earth’s surface when the grav¬
itational field is 10 N/kg is:
(a) 20 N (b) 0.2 N
(c) 20 kg (d) 5 N
4. The force required to accelerate a loaded
barrow of 80 kg mass up to 0.2 m/s 2 on
friction-less bearings is:
(a) 400 N (b) 3.2 N
(c) 0.0025 N (d) 16 N
5. A bucket of cement of mass 30 kg is tied to
the end of a rope connected to a hoist. If the
gravitational field g = 10 N/kg, the tension
in the rope when the bucket is suspended
but stationary is:
(a) 300 N (b) 3 N
(c) 300 kg (d) 0.67 N
A man of mass 75 kg is standing in a lift of mass
500 kg. Use this data to determine the answers to
questions 6 to 9. Take g as 10 m/s 2 .
6. The tension in a cable when the lift is mov¬
ing at a constant speed vertically upward is:
(a) 4250 N (b) 5750 N
(c) 4600 N (d) 6900 N
7. The tension in the cable supporting the lift
when the lift is moving at a constant speed
vertically downwards is:
(a) 4250 N (b) 5750 N
(c) 4600 N (d) 6900 N
Part Three
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196 Mechanical Engineering Principles
8. The reaction force between the man and the
floor of the lift when the lift is travelling at
a constant speed vertically upwards is:
(a) 750 N (b) 900 N
(c) 600 N (d) 475 N
9. The reaction force between the man and the
floor of the lift when the lift is travelling at
a constant speed vertically downwards is:
(a) 750 N (b) 900 N
(c) 600 N (d) 475 N
A ball of mass 0.5 kg is tied to a thread and rotated
at a constant angular velocity of 10 rad/s in a circle
of radius 1 m. Use this data to determine the an¬
swers to questions 10 and 11.
The
(a)
(c)
centripetal acceleration is:
. 100
50 m/s 2 (b) —— m/s 2
2 k
— m/s 2 (d) 100 m/s 2
2k
11
12 .
13,
The tension in the thread is:
(a) 25 N ~ ' 50
25
(c)
2k
N
(b) — N
2 K
(d) 50 N
Which of the following statements is false?
(a) An externally applied force is needed to
change the direction of a moving object.
(b) For every force, there is an equal and
opposite reaction force.
(c) A body travelling at a constant velocity
in a circle has no acceleration.
(d) Centripetal acceleration acts towards
the centre of the circle of motion.
An angular acceleration of 10 rad/s 2 occurs
when a circular disc of mass moment of
inertia of 0.5 kg m 2 is subjected to a torque.
The value of the torque is:
(a) 25 N m
(c) 20 Nm
(b) 5 N m
(d) 0.05 Nm
For fully worked solutions to each of the problems in Practice Exercises 83 to 87 in this chapter,
go to the website:
www.routledge.com/cw/bird
Chapter 16
Why it is important to understand: Work, energy and power
This chapter commences by defining work, power and energy. It also provides the mid-ordinate rule,
together with an explanation on how to apply it to calculate the areas of irregular figures, such as the
areas of ships’ water planes. It can also be used for calculating the work done in a force-displacement or
similar relationship, which may result in the form of an irregular two-dimensional shape. This chapter
is fundamental to the study and application of dynamics to practical problems. The importance of this
chapter is particularly for designing motor vehicle engines.
At the end of this chapter you should be able to:
define work and state its unit
perform simple calculations on work done
appreciate that the area under a force/distance graph gives work done
perform calculations on a force/distance graph to determine work done
define energy and state its unit
state several forms of energy
state the principle of conservation of energy and give examples of conversions
define and calculate efficiency of systems
define power and state its unit
understand that power = force x velocity
perform calculations involving power, work done, energy and efficiency
define potential energy
perform calculations involving potential energy = mgh
define kinetic energy ^
perform calculations involving kinetic energy = — mv 2
distinguish between elastic and inelastic collisions
perform calculations involving kinetic energy in rotation = — Ico 2
16.1 Work
If a body moves as a result of a force being applied to
it, the force is said to do work on the body. The amount
of work done is the product of the applied force and the
distance, i.e.
work done = force x distance moved in the
direction of the force
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
198 Mechanical Engineering Principles
The unit of work is the joule*, J, which is defined as
the amount of work done when a force of 1 newton acts
for a distance of 1 m in the direction of the force. Thus,
1 J = lNm
If a graph is plotted of experimental values of force
(on the vertical axis) against distance moved (on
the horizontal axis) a force/distance graph or work
diagram is produced. The area under the graph
represents the work done.
For example, a constant force of 20 N used to raise
a load a height of 8 m may be represented on a force/
distance graph as shown in Figure 16.1. The area
20 -
z
§ 10 -
LL
0
Figure 16.1
_i_
4 8
Distance/m
* lames Prescott Joule (24 December 1818-11 October 1889)
studied the nature of heat and discovered its relationship to
mechanical work. The SI derived unit of energy, the joule, is
named after him.
To find out more go to www.routledge.com/cw/bird
under the graph shown shaded represents the work
done. Hence
work done = 20 N x 8 m = 160 J
Similarly, a spring extended by 20 mm by a force of
500 N may be represented by the work diagram shown
in Figure 16.2, where
Figure 16.2
1
work done = shaded area = — x base x height
= 2 x (20 x 1(H) m x 500 N = 5 J
It is shown in Chapter 15 that force = mass x accelera¬
tion, and that if an object is dropped from a height it has a
constant acceleration of around 9.81 m/s 2 . Thus if a mass
of 8 kg is lifted vertically 4 m, the work done is given by:
work done = force x distance
= (mass x acceleration) x distance
= (8 x 9.81) x 4 = 313.92 J
The work done by a variable force may be found by deter¬
mining the area enclosed by the force/distance graph using
an approximate method such as the mid-ordinate rule.
Figure 16.3
To determine the area ABCD of Figure 16.3 using the
mid-ordinate rule:
(i) Divide base AD into any number of equal inter¬
vals, each of width d (the greater the number of
intervals, the greater the precision)
(ii) Erect ordinates in the middle of each interval
(shown by broken lines in Figure 16.3)
(iii) Accurately measure ordinates y l ,y 2 , y 2 , etc.
(iv) Area ABCD = d(y x + y 2 +y 3 + y 4 + y 5 + y 6 )
In general, the mid-ordinate rule states:
Area = (width of interval) (sum of mid-ordinates)
Work, energy and power 199
Calculate the work done when a
force of 40 N pushes an object a distance of 500 m
in the same direction as the force.
Work done = force x distance moved in the direction
of the force
= 40 N x 500 m
= 20000 J (since 1 J = 1 N m)
i.e. work done = 20 kJ
Calculate the work done when a
mass is lifted vertically by a crane to a height of
5 m, the force required to lift the mass being 98 N.
When work is done in lifting then:
work done = (weight of the body)
x (vertical distance moved)
Weight is the downward force due to the mass of an
object. Hence
work done = 98 N x 5 m = 490 J
A motor supplies a constant force of
1 kN which is used to move a load a distance of 5 m.
The force is then changed to a constant 500 N and
the load is moved a further 15 m. Draw the force/
distance graph for the operation and from the graph
determine the work done by the motor.
The force/distance graph or work diagram is shown in
Figure 16.4. Between points A and B a constant force of
1000 N moves the load 5 m; between points C and D a
constant force of 500 N moves the load from 5 m to 20 m.
A R
1000
500
C
D
E
i
G
0
c
10 15 20
Distance/m
Figure 16.4
Total work done = area under the force/distance graph
= area ABFE + area CDGF
= (1000 N x 5 m) + (500 N x 15 m)
= 5000 J + 7500 J = 12500 J = 12.5 kJ
A spring, initially in a relaxed state, is
extended by 100 mm. Determine the work done by
using a work diagram if the spring requires a force
of 0.6 N per mm of stretch.
Force required for a 100 mm extension =100 mm
x 0.6 N/mm = 60 N. Figure 16.5 shows the force/
extension graph or work diagram representing the
increase in extension in proportion to the force, as the
force is increased from 0 to 60 N. The work done is
the area under the graph, hence
work done =
— x base x height
]r x 100 mm x 60 N
jL*
x 100 x l(T 3 mx60N = 3 J
(Alternatively, average force during extension
(60 - 0)
= - = 30 N
and total extension =100 mm = 0.1 m, hence
work done = average force x extension
= 30 N x 0.1 m = 3 J)
A spring requires a force of 10 N to
cause an extension of 50 mm. Determine the work
done in extending the spring (a) from zero to 30 mm,
and (b) from 30 mm to 50 mm.
Figure 16.6 shows the force/extension graph for the
spring.
Figure 16.6
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200 Mechanical Engineering Principles
Work done in extending the spring from zero to
30 mm is given by area^O of Figure 16.6,
i.e. work done = ^ x base x height
= I x 30 x l(T 3 m x 6 N
= 90 x 10 3 J = 0.09 J
(b) Work done in extending the spring from 30 mm to
50 mm is given by area ABCE of Figure 16.6, i.e.
work done = area ABCD + area ADE
= (20 x l(T 3 m X 6 N) + T (20 x l(T 3 m)(4 N)
= 0.12 J + 0.04 J = 0.16 J
Calculate the work done when a mass
of 20 kg is lifted vertically through a distance of
5.0 m. Assume that the acceleration due to gravity
is 9.81 m/s 2 .
The force to be overcome when lifting a mass of 20 kg
vertically upwards is mg,
i.e. 20 x 9.81 = 196.2 N (see Chapter 15).
Work done = force x distance = 196.2 x 5.0 = 981 J
Water is pumped vertically upwards
through a distance of 50.0 m and the work done is
294.3 kJ. Determine the number of litres of water
pumped. (1 litre of water has a mass of 1 kg).
Work done = force x distance
i.e. 294300 = force x 50.0
from which, force =
294300
50.0
= 5886 N
The force to be overcome when lifting a mass m kg
vertically upwards is mg,
i.e. (m x 9.81) N (see Chapter 15).
Thus, 5886 = mx9.81, from which
mass, m = ^ = 600 kg.
Since 1 litre of water has a mass of 1 kg, 600 litres of
water are pumped.
The force on a cutting tool of a
shaping machine varies over the length of cut as
follows:
Distance (mm)
0 20
40 60 80
100
Force (kN)
60 72
65 53 44
50
Determine the work done as the tool moves
through a distance of 100 mm.
The force/distance graph for the given data is shown in
Figure 16.7. The work done is given by the area under
the graph; the area may be determined by an approxi¬
mate method. Using the mid-ordinate rule, with each
strip of width 20 mm, mid-ordinates y v y 2 ^y^y 4
are erected as shown, and each is measured.
Figure 16.7
Area under curve = (width of each strip) (sum of
mid-ordinate values)
= (20)(69 + 69.5 + 59 + 48 + 45.5)
= (20)(291)
= 5820 kN mm = 5820 Nm
= 5820 J
Hence the work done as the tool moves through
100 mm is 5.82 kJ
Work, energy and power 201
Now try the following Practice Exercise
Practice Exercise 88 Further problems
on work
1. Determine the work done when a force of
50 N pushes an object 1.5 km in the same
direction as the force. [75 kJ]
2. Calculate the work done when a mass of
weight 200 N is lifted vertically by a crane
to a height of 100 m. [20 kJ]
3. A motor supplies a constant force of 2 kN to
move a load 10 m. The force is then changed
to a constant 1.5 kN and the load is moved a
further 20 m. Draw the force/distance graph
for the complete operation, and, from the
graph, determine the total work done by the
motor. [50 kJ]
4. A spring, initially relaxed, is extended
80 mm. Draw a work diagram and hence de¬
termine the work done if the spring requires
a force of 0.5 N/mm of stretch. [1.6 J]
5. A spring requires a force of 50 N to cause an
extension of 100 mm. Determine the work
done in extending the spring (a) from 0 to
100 mm, and (b) from 40 mm to 100 mm.
[(a) 2.5 J (b) 2.1 J]
6. The resistance to a cutting tool varies dur¬
ing the cutting stroke of 800 mm as fol¬
lows: (i) the resistance increases uniformly
from an initial 5000 N to 10000 N as the tool
moves 500 mm, and (ii) the resistance falls
uniformly from 10000 N to 6000 N as the
tool moves 300 mm. Draw the work diagram
and calculate the work done in one cutting
stroke. [6.15 kJ]
16.2 Energy
Energy is the capacity, or ability, to do work. The unit
of energy is the joule, the same as for work. Energy is
expended when work is done. There are several forms
of energy and these include:
(i) Mechanical energy
(ii) Heat or thermal energy
(iii) Electrical energy
(iv) Chemical energy
(v) Nuclear energy
(vi) Light energy
(vii) Sound energy
Energy may be converted from one form to another.
The principle of conservation of energy states that
the total amount of energy remains the same in such
conversions, i.e. energy cannot be created or destroyed.
Some examples of energy conversions include:
(i) Mechanical energy is converted to electrical
energy by a generator
(ii) Electrical energy is converted to mechanical
energy by a motor
(iii) Heat energy is converted to mechanical energy
by a steam engine
(iv) Mechanical energy is converted to heat energy
by friction
(v) Heat energy is converted to electrical energy by
a solar cell
(vi) Electrical energy is converted to heat energy by
an electric fire
(vii) Heat energy is converted to chemical energy by
living plants
(viii) Chemical energy is converted to heat energy by
burning fuels
(ix) Heat energy is converted to electrical energy by
a thermocouple
(x) Chemical energy is converted to electrical
energy by batteries
(xi) Electrical energy is converted to light energy by
a light bulb
(xii) Sound energy is converted to electrical energy
by a microphone
(xiii) Electrical energy is converted to chemical
energy by electrolysis.
Efficiency is defined as the ratio of the useful output
energy to the input energy. The symbol for efficiency is
rj (Greek letter eta). Hence
useful output energy
pp • * oi/
efficiency, ti =-;---
input energy
Efficiency has no units and is often stated as a percent¬
age. A perfect machine would have an efficiency of
100%. However, all machines have an efficiency lower
than this due to friction and other losses. Thus, if the
input energy to a motor is 1000 J and the output energy
is 800 J then the efficiency is x 100% = 80%
A machine exerts a force of 200 N
in lifting a mass through a height of 6 m. If 2 kJ of
energy are supplied to it, what is the efficiency of
the machine?
Part Three
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202 Mechanical Engineering Principles
Work done in lifting mass = force x distance moved
= weight of body x distance moved
= 200 N x 6 m = 1200 J
= useful energy output
Energy input = 2 kJ = 2000 J
useful output energy
Efficiency, #/
input energy
1200 = 0.6 or 60%
2000
Calculate the useful output energy
of an electric motor which is 70% efficient if it
uses 600 J of electrical energy.
Efficiency, rj =
useful output energy
input energy
thus
70 = output energy
100 600J
from which,
output energy =
70
100
x 600 = 420 J
4 kJ of energy are supplied to a
machine used for lifting a mass. The force required
is 800 N. If the machine has an efficiency of 50%,
to what height will it lift the mass?
Efficiency, // =
i.e.
50
Too
useful output energy
input energy
output energy
4000 J
from which, output energy = x 4000 = 2000 J
Work done = force x distance moved
hence 2000 J = 800 N x height
2000 J
from which, height = ^qq^ ' = 2.5 m
A hoist exerts a force of 500 N
in raising a load through a height of 20 m. The
efficiency of the hoist gears is 75% and the
efficiency of the motor is 80%. Calculate the
input energy to the hoist.
The hoist system is shown diagrammatically in Figure
16.8.
Output energy = work done = force x distance
- 500 Nx 20 m= 10000 J
. output energy
For the gearing, efficiency = --
input energy
75 10000
i.e. -= --
100 input energy
from which, the input energy to the gears
= 10000 x ^ = 13333 J
The input energy to the gears is the same as the output
energy of the motor. Thus, for the motor,
_ . output energy
efficiency = --
input energy
80 _ 13333
100 input energy
Hence, input energy to the hoist = 13333 x
100
"80"
= 16667 J = 16.67 kJ
Now try the following Practice Exercise
Practice Exercise 89 Further problems on
energy
1. A machine lifts a mass of weight 490.5 N
through a height of 12 m when 7.85 kJ of
energy is supplied to it. Determine the
efficiency of the machine. [75%]
2. Determine the output energy of an electric
motor which is 60% efficient if it uses 2 kJ
of electrical energy. [1.2 kJ]
3. A machine that is used for lifting a particular
mass is supplied with 5 kJ of energy. If the
machine has an efficiency of 65% and exerts
a force of 812.5 N to what height will it lift
the mass? [4 m]
4. A load is hoisted 42 m and requires a force
of 100 N. The efficiency of the hoist gear is
60% and that of the motor is 70%. Deter¬
mine the input energy to the hoist. [ 10 kJ]
Figure 16.8 at which energy is converted from one form to another.
Work, energy and power 203
„ „ energy used „ work done
Power P= — —- or P = — ——
time taken time taken
The unit of power is the watt*, W, where 1 watt is
equal to 1 joule per second. The watt is a small unit for
many purposes and a larger unit called the kilowatt,
kW, is used, where 1 kW = 1000 W.
The power output of a motor, which does 120 kJ of
work in 30 s, is thus given by
„ work done
Power = —-:—
time taken
from which, work done = powerx time = 8000 Wx30 s
= 240000 J = 240 kJ
Calculate the power required to lift
a mass through a height of 10 m in 20 s if the force
required is 3924 N.
120kJ
30s
= 4 kW
Work done = force x distance moved
= 3924 N x 10 m = 39240 J
Since work done = force x distance, then
Power = work done = force x distance
time taken time taken
= force x distance
time taken
However, distance = velocity
time taken
Hence, power = force x velocity
The output power of a motor is
8 kW. How much work does it do in 30 s?
* lames Watt (19 January 1736-25 August 1819) was a Scot¬
tish inventor and mechanical engineer. The watt is named after
him — the unit of power incorporated in the International Sys¬
tem of Units CSV).
To find out more go to www.routledge.com/cw/bird
_ work done
Power = —- T —
time taken
39240J
20s
- 1962 W or 1.962 kW
10 kJ of work is done by a force in
moving a body uniformly through 125 m in 50 s.
Determine (a) the value of the force, and (b) the
power.
(a) Work done = force x distance
hence 10000 J = force x 125 m
from which, force
10000J
125 m
= 80 N
_ work done
Power = —--—
time taken
10000J
50 s
= 200 W
A car hauls a trailer at 90 km/h
when exerting a steady pull of 600 N. Calculate
(a) the work done in 30 minutes and (b) the power
required.
Work done = force x distance moved.
The distance moved in 30 min, i.e.
90 km/h = 45 km.
Hence, work done = 600 N x 45000 m
at
Power required =
= 27000 kJ or 27 MJ
work done 27 x 10 6 J
time taken 30 x 60s
= 15000 W or 15 kW
To what height will a mass of
weight 981 N be raised in 40 s by a machine using
a power of 2 kW?
Work done = force x distance. Hence,
work done = 981 N x height.
Part Three
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204 Mechanical Engineering Principles
n work done r- , • ,
Power = —--—, irom which,
time taken
work done = power x time taken
= 2000 W x 40 s = 80000 J
Hence, 80000 = 981 N x height, from which,
80000J
height = - 981N - = 81.55 m
A planing machine has a cutting
stroke of 2 m and the stroke takes 4 seconds. If
the constant resistance to the cutting tool is 900 N,
calculate for each cutting stroke (a) the power
consumed at the tool point, and (b) the power input
to the system if the efficiency of the system is 75%.
(a) Work done in each cutting stroke
= force x distance = 900 N x 2 m = 1800 J
Power consumed at tool point
work done 1800J
time taken
4s
= 450 W
(b) Efficiency
output energy output power
Hence,
input energy
75 450
100 input power
100
input power
from which,
input power = 450 x = 600 W
An electric motor provides power to
a winding machine. The input power to the motor is
2.5 kW and the overall efficiency is 60%. Calculate
(a) the output power of the machine, (b) the rate at
which it can raise a 300 kg load vertically upwards.
(a) Efficiency, rj
power output
power input
i.e.
60 _ power output
100 2500
from which, power output
60
x 2500
(b)
100
= 1500 W or 1.5 kW
Power output = force x velocity, from which,
, power output
velocity = ---—
force
Force acting on the 300 kg load due to gravity
= 300 kg x 9.81 m/s 2 = 2943 N
1500
Hence, velocity
2943
= 0.510 m/s or 510 mm/s
A lorry is travelling at a constant
velocity of 72 km/h. The force resisting motion is
800 N. Calculate the tractive power necessary to
keep the lorry moving at this speed.
Power = force x velocity.
The force necessary to keep the lorry moving at con¬
stant speed is equal and opposite to the force resisting
motion, i.e. 800 N
72 x1000
Velocity = 72 km/h = ————- m/s = 20 m/s.
60 x 60
Hence, power = 800 N x 20 m/s
= 16000 N m/s = 16000 J/s
= 16000 W or 16 kW
Thus the tractive power needed to keep the lorry
moving at a constant speed of 72 km/h is 16 kW
The variation of tractive force with
distance for a vehicle which is accelerating from
rest is:
Force (kN) 8.0 7.4 5.8 4.5 3.7 3.0
Distance (m) 0 10 20 30 40 50
Determine the average power necessary if the time
taken to travel the 50 m from rest is 25 s.
The force/distance diagram is shown in Figure 16.9.
The work done is determined from the area under the
curve. Using the mid-ordinate rule with five intervals
gives:
0 10 20 30 40 50
Distance (m)
Figure 16.9
Work, energy and power 205
area = (width of interval)(sum of mid-ordinate)
= (io )|> 1 + y 2 + y$ + )’4 + j 5 ]
= (10)[7.8 + 6.6 + 5.1 + 4.0 + 3.3]
= (10)[26.8] = 268 kN m
i.e. work done = 268 kJ
. work done 268000J
Average power = —--— = ——-
time taken 25 s
= 10720 W or 10.72 kW
Now try the following Practice Exercise
Practice Exercise 90 Further problems on
power
1. The output power of a motor is 10 kW.
How much work does it do in 1 minute?
[600 kJ]
2. Determine the power required to lift a load
through a height of 20 m in 12.5 s if the
force required is 2.5 kN. [4 kW]
3. 25 kJ of work is done by a force in moving
an object uniformly through 50 m in 40 s.
Calculate (a) the value of the force, and
(b) the power. [(a) 500 N (b) 625 W]
4. A car towing another at 54 km/h exerts a
steady pull of 800 N. Determine (a) the
work done in hr, and (b) the power
required. [(a) 10.8 MJ (b) 12 kW]
5. To what height will a mass of weight 500 N
be raised in 20 s by a motor using 4 kW of
power? [160 m]
6. The output power of a motor is 10 kW.
Determine (a) the work done by the motor
in 2 hours, and (b) the energy used by the
motor if it is 72% efficient.
[(a) 72 MJ (b) 100 MJ]
7. A car is travelling at a constant speed
of 81 km/h. The frictional resistance to
motion is 0.60 kN. Determine the power
required to keep the car moving at this
speed. [13.5 kW]
8. A constant force of 2.0 kN is required to
move the table of a shaping machine when
a cut is being made. Determine the power
required if the stroke of 1.2 m is completed
in 5.0 s. [480 W]
9. The variation of force with distance for a
vehicle that is decelerating is as follows:
Distance
600
500
400
300
200
100
0
(m)
Force
(kN)
24
20
16
12
8
4
0
If the vehicle covers the 600 m in 1.2 min¬
utes, find the power needed to bring the
vehicle to rest. [ 100 kW]
10. A cylindrical bar of steel is turned in a
lathe. The tangential cutting force on the
tool is 0.5 kN and the cutting speed is
180 mm/s. Determine the power absorbed
in cutting the steel. [90 W]
16.4 Potential and kinetic energy
Mechanical engineering is concerned principally with
two kinds of energy: potential energy and kinetic
energy.
Potential energy is energy due to the position of
the body. The force exerted on a mass of m kg is
mg N (where g = 9.81 m/s * 1 2 3 4 5 6 7 8 , the acceleration due to
gravity). When the mass is lifted vertically through a
height h m above some datum level, the work done is
given by:
force x distance = ( mg)(h ) J
This work done is stored as potential energy in the
mass.
Hence, potential energy = mgh joules
(the potential energy at the datum level being taken as
zero).
Kinetic energy is the energy due to the motion of
a body. Suppose a force F acts on an object of mass
m originally at rest (i.e. u = 0) and accelerates it to a
velocity v in a distance s :
work done = force x distance
= Fs = ( ma)(s) (if no energy is lost)
where a is the acceleration
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206 Mechanical Engineering Principles
Since
v 2 = u 2 + las (see Chapter 13)
,2
,2 _
and u = 0, = las , from which a
hence, work done = (ma)(s) = (m)
v
Is
2 ^
V 7
(s) = \ mv 1
This energy is called the kinetic energy of the mass m ,
1 ?
i.e. kinetic energy = —mv L joules
As stated in Section 16.2, energy may be converted
from one form to another. The principle of conserva-
ergy states that the total amount of energy
remains the same in such conversions, i.e. energy can¬
not be created or destroyed.
In mechanics, the potential energy possessed by a
body is frequently converted into kinetic energy, and
vice versa. When a mass is falling freely, its poten¬
tial energy decreases as it loses height, and its kinetic
energy increases as its velocity increases. Ignoring air
frictional losses, at all times:
Potential energy + kinetic energy = a constant
If friction is present, then work is done overcoming
the resistance due to friction and this is dissipated as
heat. Then,
Initial energy = final energy
+ work done overcoming frictional resistance
Kinetic energy is not always conserved in collisions.
Collisions in which kinetic energy is conserved (i.e.
stays the same) are called elastic collisions, and those in
which it is not conserved are termed inelastic collisions.
A car of mass 800 kg is climbing
an incline at 10° to the horizontal. Determine the
increase in potential energy of the car as it moves a
distance of 50 m up the incline.
With reference to Figure 16.10,
sin 10° =
opposite
hypotenuse
h
50
from which, h = 50 sin 10° = 8.682 m
Figure 16.10
Hence, increase in potential energy = mgh
= 800 kg x 9.81 m/s 2 x 8.682 m
= 68140 J or 68.14 kJ
At the instant of striking, a hammer
of mass 30 kg has a velocity of 15 m/s. Determine
the kinetic energy in the hammer.
Kinetic energy =
\ (30 kg)(15 m/s) 2
i.e. kinetic energy in hammer = 3375 J or 3.375 kJ
A lorry having a mass of 1.5 t is
travelling along a level road at 72 km/h. When the
brakes are applied, the speed decreases to 18 km/h.
Determine how much the kinetic energy of the lorry
is reduced.
Initial velocity of lorry,
Vj = 72 km/h = 72 ^ x 1000 x -T-
1 h km 3600 s
72
3.6
final velocity of lorry,
18
= 20 m/s,
v
= 5 m/s and mass of lorry,
2 3.6
m = 1.5 t = 1500 kg
Initial kinetic energy of the lorry = — m Vj
= 1 (1500)(20) ;
= 300 kJ
1
Final kinetic energy of the lorry = m v 2
1
- - (1500)(5) 2
- 18.75 kJ
Hence, the change in kinetic energy = 300 - 18.75
= 281.25 kJ
(Part of this reduction in kinetic energy is converted into
heat energy in the brakes of the lorry and is hence dis¬
sipated in overcoming frictional forces and air friction.)
A canister containing a meteorology
balloon of mass 4 kg is fired vertically upwards
from a gun with an initial velocity of 400 m/s.
Neglecting the air resistance, calculate (a) its initial
kinetic energy (b) its velocity at a height of 1 km
(c) the maximum height reached.
Initial kinetic energy =
2 mV
I (4)(400) 2 = 320 kJ
Work, energy and power 207
(b) At a height of 1 km, potential energy
- mgh = 4 x 9.81 x 1000 - 39.24 kJ
By the principle of conservation of energy:
potential energy + kinetic energy at 1 km = initial
kinetic energy.
Hence 39240 + 3 mv 2 = 320000
from which, \ (4)v 2 = 320000 - 39240 = 280760
Hence,
'lx 280760 A
V
V
/
= 374.7 m/s
i.e. the velocity of the canister at a height of
1 km is 374.7 m/s
(c) At the maximum height, the velocity of the can¬
ister is zero and all the kinetic energy has been
converted into potential energy. Hence,
potential energy = initial kinetic energy
= 320000 J (from part (a))
Then,
from which, height h
320000 = mgh = (4)(9.81)(h)
320000
(4)(9.81)
=8155 m
i.e. the maximum height reached is 8155 m or
8.155 km
A pile-driver of mass 500 kg falls
freely through a height of 1.5 m on to a pile of mass
200 kg. Determine the velocity with which the
driver hits the pile. If, at impact, 3 kJ of energy are
lost due to heat and sound, the remaining energy
being possessed by the pile and driver as they
are driven together into the ground a distance of
200 mm, determine (a) the common velocity
immediately after impact (b) the average resistance
of the ground.
The potential energy of the pile-driver is converted into
kinetic energy.
Thus potential energy = kinetic energy, i.e.
mgh = —mv 2
from which, velocity v = -yjlgh
= -7(2X9.81X1.5) = 5.42 m/s.
Hence, the pile-driver hits the pile at a velocity of
5.42 m/s
(a) Before impact, kinetic energy of pile driver
1
2 mV
2 -
1
7(500)(5.42) 2 = 7.34 kJ
Kinetic energy after impact = 7.34-3 = 4.34 kJ.
Thus the pile-driver and pile together have a
mass of 500 + 200 = 700 kg and possess kinetic
energy of 4.34 kJ
Hence, 4.34 x 10 3 = = ]r (700)v 2
jL* ^
from which, velocity v
W 2 x 4.34 x 10 3 ^
700
(b)
= 3.52 m/s
Thus, the common velocity after impact is
3.52 m/s.
The kinetic energy after impact is absorbed in
overcoming the resistance of the ground, in a dis¬
tance of 200 mm.
Kinetic energy = work done
= resistance x distance
i.e. 4.34 x 10 3 = resistance x 0.200
4.34 xlO 3
from which, resistance = —q^qq— = 21700 N
Hence, the average resistance of the ground is
21.7 kN
A car of mass 600 kg reduces speed
from 90 km/h to 54 km/h in 15 s. Determine the
braking power required to give this change of
speed.
1 ? 1 9
Change in kinetic energy of car = — m Vj - —mv 2 z
where m = mass of car = 600 kg,
Vj = initial velocity = 90 km/h
= ~~ m/s = 25 m/s,
3.6
and v 2 = final velocity = 54 km/h
54
3.6
m/s =15 m/s.
Hence, change in kinetic energy = — m(v 2 - v 2 2 )
= ^(600)(25 2 - 15 2 )
= 120000 J
Part Three
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208 Mechanical Engineering Principles
Braking power
change in energy
time taken
120000J
15s
= 8000 W or 8 kW
Now try the following Practice Exercise
Practice Exercise 91 Further problems on
potential and kinetic
energy
(Assume the acceleration due to gravity,
g = 9.81 m/s 2 )
1. An object of mass 400 g is thrown verti¬
cally upwards and its maximum increase
in potential energy is 32.6 J. Determine
the maximum height reached, neglecting air
resistance. [8.31m]
2. A ball bearing of mass 100 g rolls down from
the top of a chute of length 400 m inclined at
an angle of 30° to the horizontal. Determine
the decrease in potential energy of the ball
bearing as it reaches the bottom of the chute.
[196.2 J]
3. A vehicle of mass 800 kg is travelling at
54 km/h when its brakes are applied. Find
the kinetic energy lost when the car comes to
rest. [90 kJ]
4. A body of mass 15 kg has its speed reduced
from 30 km/h to 18 km/h in 4.0 s. Calculate
the power required to effect this change of
speed. [83.33 W]
5. Supplies of mass 300 kg are dropped from
a helicopter flying at an altitude of 60 m.
Determine the potential energy of the sup¬
plies relative to the ground at the instant of
release, and its kinetic energy as it strikes the
ground. [176.6 kJ, 176.6 kJ]
6. A shell of mass 10 kg is fired vertically
upwards with an initial velocity of 200 m/s.
Determine its initial kinetic energy and the
maximum height reached, correct to the
nearest metre, neglecting air resistance.
[200 kJ, 2039 m]
7. The potential energy of a mass is increased
by 20.0 kJ when it is lifted vertically through
a height of 25.0 m. It is now released and
allowed to fall freely. Neglecting air resis¬
tance, find its kinetic energy and its velocity
after it has fallen 10.0 m. [8 kJ, 14.0 m/s]
8. A pile-driver of mass 400 kg falls freely
through a height of 1.2 m on to a pile of mass
150 kg. Determine the velocity with which
the driver hits the pile. If, at impact, 2.5 kJ of
energy are lost due to heat and sound, the re¬
maining energy being possessed by the pile
and driver as they are driven together into
the ground a distance of 150 mm, determine
(a) the common velocity after impact (b) the
average resistance of the ground.
[4.85 m/s (a) 2.83 m/s (b) 14.70 kN]
16.5 Kinetic energy of rotation
When linear motion takes place,
kinetic energy = 5/
m 2
v
but when rotational motion takes place,
kinetic energy = -m(cor) 2
Since a; is a constant, kinetic energy = co 2 —X,
mr
But
mr 2 = 1
Therefore, kinetic energy (in rotation)
= ^ Icq 2 joules
where / = the mass moment of inertia about the
point of rotation
and co = angular velocity.
Calculate the kinetic energy of a solid
flat disc of diameter 0.5 m and of a uniform thickness
of 0.1 m, rotating about its centre at 40 rpm. Take the
density of the material as 7860 kg/m 3 .
rad rev
Angular velocity, co = 2n - x 40 —— x
rev mm
= 4.189 rad/s
1 min
60s
From Table 15.1, page 194,
T d2 Rl
I = P X 7tR Z X t X —
Work, energy and power 209
= 7860 -^-X7rx 0.25 * 1 2 m 2 x0.1 m X °’ 25 ^
m 3 4 5 6 7 2
i.e. 7=4.823 kg m 2
Hence,
kinetic energy = —Icq 2
= - X 4.823kgm 2 x (4.189) 2 \
2 s 2
= 42.32 J
Now try the following Practice Exercises
Practice Exercise 92 Further problems
on kinetic energy of
rotation
1. Calculate the kinetic energy of a solid flat
disc of diameter 0.6 m and of uniform thick¬
ness of 0.1 m rotating about its centre at
50 rpm. Take the density of the disc material
as 7860 kg/m 3 . [137.1 J]
2. If the disc of Problem 1 had a hole in its
centre of 0.2 m diameter, what would be its
kinetic energy? [135.4 J]
3. If an annulus of external diameter 0.4 m and
internal diameter 0.2 m were rotated about
its centre at 100 rpm, what would be its
kinetic energy? Assume the uniform
thickness of the annulus is 0.08 m and the
density of the material is 7860 kg/m 3 .
[81.2 J]
Practice Exercise 93 Short-answer
questions on work,
energy and power
1. Define work in terms of force applied and
distance moved.
2. Define energy, and state its unit.
3. Define the joule.
4. The area under a force/distance graph rep¬
resents .
5. Name five forms of energy.
6. State the principle of conservation of energy.
7. Give two examples of conversion of heat
energy to other forms of energy.
8. Give two examples of conversion of
electrical energy to other forms of energy.
9. Give two examples of conversion of
chemical energy to other forms of energy.
10. Give two examples of conversion of
mechanical energy to other forms of energy.
11. (a) Define efficiency in terms of energy
input and energy output.
(b) State the symbol used for efficiency.
12. Define power and state its unit.
13. Define potential energy.
14. The change in potential energy of a body of
mass m kg when lifted vertically upwards
to a height h m is given by.
15. What is kinetic energy?
16. The kinetic energy of a body of mass m kg
and moving at a velocity of v m/s is given
by.
17. State the principle of conservation of energy.
18. Distinguish between elastic and inelastic
collisions.
19. The kinetic energy of rotation of a body of
moment of inertia 7 kg m 2 and moving at an
angular velocity of co rad/s is given by.
Practice Exercise 94 Multiple-choice
questions on work,
energy and power
(Answers on page 336)
1. State which of the following is incorrect:
(a) 1 W = 1 J/s
(b) 1 J = 1 N/m
. . output energy
(c) 77 = -;- - - —
input energy
(d) energy = power x time
2. An object is lifted 2000 mm by a crane. If the
force required is 100 N, the work done is:
(a) T N m (b) 200 kN m
(c) 200 Nm (d) 20 J
3. A motor having an efficiency of 0.8 uses
800 J of electrical energy. The output
energy of the motor is:
Part Three
Part Three
210 Mechanical Engineering Principles
(a) 800 J (b) 1000 J
(c) 640 J (d) 6.4 J
4. 6 kJ of work is done by a force in mov¬
ing an object uniformly through 120 m
in 1 minute. The force applied is:
(a) 50 N (b) 20 N
(c) 720 N (d) 12 N
5. For the object in question 4, the power
developed is:
(a) 6 kW (b) 12 kW
(c) 5/6 W (d)0.1kW
6. Which of the following statements is false?
(a) The unit of energy and work is the same.
(b) The area under a force/distance graph
gives the work done.
(c) Electrical energy is converted to me¬
chanical energy by a generator.
(d) Efficiency is the ratio of the useful out¬
put energy to the input energy.
7. A machine using a power of 1 kW requires
a force of 100 N to raise a mass in 10 s. The
height the mass is raised in this time is:
(a) 100 m (b) 1km
(c) 10 m (d) 1 m
8. A force/extension graph for a spring is
shown in Figure 16.11. Which of the
following statements is false?
Figure 16.11
The work done in extending the spring:
(a) from 0 to 100 mm is 5 J
(b) from 0 to 50 mm is 1.25 J
(c) from 20 mm to 60 mm is 1.6 J
(d) from 60 mm to 100 mm is 3.75 J
9. A vehicle of mass 1 tonne climbs an
incline of 30° to the horizontal. Taking the
acceleration due to gravity as 10 m/s 2 , the
increase in potential energy of the vehicle
as it moves a distance of 200 m up the
incline is:
(a) 1 kJ (b) 2 MJ
(c) 1 MJ (d) 2 kJ
10. A bullet of mass 100 g is fired from a gun
with an initial velocity of 360 km/h. Ne¬
glecting air resistance, the initial kinetic
energy possessed by the bullet is:
(a) 6.48 kJ (b) 500 J
(c) 500 kJ (d) 6.48 MJ
11. A small motor requires 50 W of electrical
power in order to produce 40 W of
mechanical energy output. The efficiency
of the motor is:
(a) 10% (b) 80%
(c) 40% (d) 90%
12. A load is lifted 4000 mm by a crane. If the
force required to lift the mass is 100 N, the
work done is:
(a) 400 J (b) 40 Nm
(c) 25 J (d) 400 kJ
13. A machine exerts a force of 100 N in lift¬
ing a mass through a height of 5 m. If 1 kJ
of energy is supplied, the efficiency of the
machine is:
(a) 10% (b) 20%
(c) 100% (d) 50%
14. At the instant of striking an object, a ham¬
mer of mass 40 kg has a velocity of 10 m/s.
The kinetic energy in the hammer is:
(a) 2 kJ (b) 1 kJ
(c) 400 J (d) 8 kJ
15. A machine which has an efficiency of
80% raises a load of 50 N through a verti¬
cal height of 10 m. The work input to the
machine is:
(a) 400 J (b) 500 J
(c) 800 J (d) 625 J
16. The formula for kinetic energy due to rota¬
tion is:
(a) mv 2 (b) mgh
co 2 9
(c) I— (d) co 2 r
For fully worked solutions to each of the problems in Practice Exercises 88 to 94 in this chapter,
go to the website:
www.routledge.com/cw/bird
Revision Test 6 Linear and angular motion, momentum and impulse, force, mass and
acceleration, work, energy and power
This Revision Test covers the material contained in Chapters 13 to 16. The marks for each question are shown in
brackets at the end of each question.
Assume, where necessary, that the acceleration due to
gravity, g = 9.81 m/s 2
1. A train is travelling at 90 km/h and has wheels of
diameter 1600 mm.
(a) Find the angular velocity of the wheels in
both rad/s and rev/min.
(b) If the speed remains constant for 2 km, de¬
termine the number of revolutions made by a
wheel, assuming no slipping occurs. (7)
2. The speed of a shaft increases uniformly from
200 revolutions per minute to 700 revolutions per
minute in 12 s. Find the angular acceleration, cor¬
rect to 3 significant figures. (5)
3. The shaft of an electric motor, initially at rest,
accelerates uniformly for 0.3 s at 20 rad/s 2 .
Determine the angle (in radians) turned through
by the shaft in this time. (4)
4. Determine the momentum of a lorry of mass 10
tonnes moving at a velocity of 81 km/h. (4)
5. A ball of mass 50 g is moving with a velocity of
4 m/s when it strikes a stationary ball of mass
25 g. The velocity of the 50 g ball after impact
is 2.5 m/s in the same direction as before im¬
pact. Determine the velocity of the 25 g ball after
impact. (7)
6. A force of 24 N acts on a body of mass 6 kg for
150 ms. Determine the change in velocity. (4)
7. The hammer of a pile-driver of mass 800 kg falls
a distance of 1.0 m on to a pile. The blow takes
place in 20 ms and the hammer does not rebound.
Determine (a) the velocity of impact (b) the
momentum lost by the hammer (c) the average
applied force exerted on the pile by the hammer.
( 8 )
8. Determine the mass of the moving head of a
machine tool if it requires a force of 1.5 N to
bring it to rest in 0.75 s from a cutting speed of
25 m/min. (5)
9. Find the weight of an object of mass 2.5 kg at
a point on the Earth’s surface where the gravita¬
tional field is 9.8 N/kg. (4)
10. A van of mass 1200 kg travels round a bend of
radius 120 m, at 54 km/h. Determine the centrip¬
etal force acting on the vehicle. (4)
11. A spring, initially in a relaxed state, is extended
by 80 mm. Determine the work done by using
a work diagram if the spring requires a force of
0.7 N per mm of stretch. (4)
12. Water is pumped vertically upwards through a
distance of 40.0 m and the work done is 176.58 kJ.
Determine the number of litres of water pumped.
(1 litre of water has a mass of 1 kg.) (4)
13. 3 kJ of energy are supplied to a machine used for
lifting a mass. The force required is 1 kN. If the
machine has an efficiency of 60%, to what height
will it lift the mass? (4)
14. When exerting a steady pull of 450 N, a lorry
travels at 80 km/h. Calculate (a) the work done in
15 minutes and (b) the power required. (4)
15. An electric motor provides power to a winding
machine. The input power to the motor is 4.0 kW
and the overall efficiency is 75%. Calculate
(a) the output power of the machine (b) the rate
at which it can raise a 509.7 kg load vertically
upwards. (4)
16. A tank of mass 4800 kg is climbing an incline at
12° to the horizontal. Determine the increase in
potential energy of the tank as it moves a distance
of 40 m up the incline. (4)
17. A car of mass 500 kg reduces speed from
108 km/h to 36 km/h in 20 s. Determine the brak¬
ing power required to give this change of speed.
(4)
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 6,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
Part Three
Chapter 17
Why it is important to understand: Friction
When a block is placed on a flat surface and sufficient force is applied to the block, the force being parallel
to the surface, the block slides across the surface. When the force is removed, motion of the block stops;
thus there is a force which resists sliding. In this chapter, both dynamic and static frictions are explained,
together with the factors that affect the size and direction of frictional forces. A low coefficient of friction
is desirable in bearings, pistons moving within cylinders and on ski runs; however, for a force being
transmitted by belt drives and braking systems, a high value of coefficient is necessary. Advantages
and disadvantages of frictional forces are discussed and calculations are performed on friction on an
inclined plane and screw jack efficiency. Knowledge of friction is of importance for the static and dynamic
behaviour of stationary and moving bodies.
At the end of this chapter you should be able to:
• understand dynamic or sliding friction
• appreciate factors which affect the size and direction of frictional forces
• define coefficient of friction, p
• perform calculations involving F = pH
• state practical applications of friction
• state advantages and disadvantages of frictional forces
• understand friction on an inclined plane
• perform calculations on friction on an inclined plane
• calculate the efficiency of a screw jack
17.1 Introduction to friction
When an object, such as a block of wood, is placed
on a floor and sufficient force is applied to the block,
the force being parallel to the floor, the block slides
across the floor. When the force is removed, motion of
the block stops; thus there is a force which resists slid¬
ing. This force is called dynamic or sliding friction. A
force may be applied to the block, which is insufficient
to move it. In this case, the force resisting motion is
called the static friction or stiction. Thus there are two
categories into which a frictional force may be split:
(i) dynamic or sliding friction force which occurs
when motion is taking place, and
(ii) static friction force which occurs before motion
takes place.
There are three factors that affect the size and direction
of frictional forces.
(i) The size of the frictional force depends on the
type of surface (a block of wood slides more eas¬
ily on a polished metal surface than on a rough
concrete surface).
(ii) The size of the frictional force depends on the
size of the force acting at right angles to the
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
Friction 213
surfaces in contact, called the normal force;
thus, if the weight of a block of wood is doubled,
the frictional force is doubled when it is sliding
on the same surface.
The direction of the frictional force is always op¬
posite to the direction of motion. Thus the frictional
force opposes motion, as shown in Figure 17.1.
Frictional force
Motion
Pulling force
Surface
Motion
Pushing force
- Frictional force
Surface
Figure 17.1
17.2 Coefficient of friction
The coefficient of friction, //, is a measure of the
amount of friction existing between two surfaces. A
low value of coefficient of friction indicates that the
force required for sliding to occur is less than the
force required when the coefficient of friction is high.
The value of the coefficient of friction is
given by:
frictional force (F)
normal force ( N )
Transposing gives: frictional force = fix normal force
i.e.
F = juN
Normal
force, N
Frictional
force, F=juN
Applied
Force, P
Figure 17.2
The direction of the forces given in this equation is as
shown in Figure 17.2.
The coefficient of friction is the ratio of a force to a
force, and hence has no units. Typical values for the
coefficient of friction when sliding is occurring, i.e. the
dynamic coefficient of friction, are:
For polished oiled metal surfaces less than 0.1
For glass on glass 0.4
For rubber on tarmac close to 1.0
The coefficient of friction (ji ) for dynamic friction
is, in general, a little less than that for static friction.
However, for dynamic friction, // increases with speed;
additionally, it is dependent on the area of the surface
in contact.
A block of steel requires a force of
10.4 N applied parallel to a steel plate to keep it
moving with constant velocity across the plate. If
the normal force between the block and the plate is
40 N, determine the dynamic coefficient of friction.
As the block is moving at constant velocity, the force
applied must be that required to overcome frictional
forces, i.e. frictional force, F = 10.4 N;
the normal force is 40 N, and since F = juN,
_ F _ 10.4
N ~ ~40~
= 0.26
i.e. the dynamic coefficient of friction is 0.26
The surface between the steel block
and plate of Problem 1 is now lubricated and the
dynamic coefficient of friction falls to 0.12. Find
the new value of force required to push the block at
a constant speed.
The normal force depends on the weight of the block
and remains unaltered at 40 N. The new value of the
dynamic coefficient of friction is 0.12 and since the
frictional force F = [iN ,
F= 0.12 x 40 = 4.8 N
The block is sliding at constant speed, thus the force
required to overcome the frictional force is also 4.8 N,
i.e. the required applied force is 4.8 N
The material of a brake is being
tested and it is found that the dynamic coefficient
of friction between the material and steel is 0.91.
Calculate the normal force when the frictional
force is 0.728 kN.
The dynamic coefficient of friction, [i = 0.91 and the
frictional force, F = 0.728 kN = 728 N
Since F = juN,
F 728
then normal force, N = — = 77—7 = 800 N
H 0.91
i.e. the normal force is 800 N
Part Three
Part Three
214 Mechanical Engineering Principles
Now try the following Practice Exercise
Practice Exercise 95 Further problems
on the coefficient of
friction
1. The coefficient of friction of a brake pad
and a steel disc is 0.82. Determine the nor¬
mal force between the pad and the disc if the
frictional force required is 1025 N.
[1250 N]
2. A force of 0.12 kN is needed to push a bale
of cloth along a chute at a constant speed.
If the normal force between the bale and
the chute is 500 N, determine the dynamic
coefficient of friction. [0.24]
3. The normal force between a belt and its driver
wheel is 750 N. If the static coefficient of
friction is 0.9 and the dynamic coefficient of
friction is 0.87, calculate
(a) the maximum force which can be trans¬
mitted, and
(b) the maximum force which can be trans¬
mitted when the belt is running at a con¬
stant speed. [(a) 675 N (b) 652.5 N]
17.3 Applications of friction
In some applications, a low coefficient of friction is de¬
sirable, for example, in bearings, pistons moving within
cylinders, on ski runs, and so on. However, for such ap¬
plications as force being transmitted by belt drives and
braking systems, a high value of coefficient is necessary.
State three advantages, and three
disadvantages of frictional forces.
Instances where frictional forces are an advantage
include:
(i) Almost all fastening devices rely on frictional
forces to keep them in place once secured, exam¬
ples being screws, nails, nuts, clips and clamps.
(ii) Satisfactory operation of brakes and clutches
rely on frictional forces being present.
(iii) In the absence of frictional forces, most accelera¬
tions along a horizontal surface are impossible; for
example, a person’s shoes just slip when walking
is attempted and the tyres of a car just rotate with
no forward motion of the car being experienced.
Disadvantages of frictional forces include:
(i) Energy is wasted in the bearings associated with
shafts, axles and gears due to heat being generated.
(ii) Wear is caused by friction, for example, in shoes,
brake lining materials and bearings.
(iii) Energy is wasted when motion through air oc¬
curs (it is much easier to cycle with the wind
rather than against it).
Discuss briefly two design implica¬
tions that arise due to frictional forces and how
lubrication may or may not help.
(i) Bearings are made of an alloy called white metal,
which has a relatively low melting point. When
the rotating shaft rubs on the white metal bearing,
heat is generated by friction, often in one spot
and the white metal may melt in this area, ren¬
dering the bearing useless. Adequate lubrication
(oil or grease) separates the shaft from the white
metal, keeps the coefficient of friction small and
prevents damage to the bearing. For very large
bearings, oil is pumped under pressure into the
bearing and the oil is used to remove the heat gen¬
erated, often passing through oil coolers before
being re-circulated. Designers should ensure that
the heat generated by friction can be dissipated.
(ii) Wheels driving belts, to transmit force from one
place to another, are used in many workshops.
The coefficient of friction between the wheel and
the belt must be high, and it may be increased
by dressing the belt with a tar-like substance.
Since frictional force is proportional to the nor¬
mal force, a slipping belt is made more efficient
by tightening it, thus increasing the normal and
hence the frictional force. Designers should
incorporate some belt tension mechanism into the
design of such a system.
Explain what is meant by the terms
(a) the limiting or static coefficient of friction, and
(b) the sliding or dynamic coefficient of friction.
(a) When an object is placed on a surface and a force
is applied to it in a direction parallel to the sur¬
face, if no movement takes place, then the applied
force is balanced exactly by the frictional force.
As the size of the applied force is increased, a
Friction 215
value is reached such that the object is just on
the point of moving. The limiting or static
coefficient of friction is given by the ratio of this
applied force to the normal force, where the nor¬
mal force is the force acting at right angles to the
surfaces in contact.
(b) Once the applied force is sufficient to overcome
the stiction or static friction, its value can be
reduced slightly and the object moves across the
surface. A particular value of the applied force
is then sufficient to keep the object moving at a
constant velocity. The sliding or dynamic coef¬
ficient of friction is the ratio of the applied force,
to maintain constant velocity, to the normal force.
17.4 Friction on an inclined plane
Angle of repose
Consider a mass m lying on an inclined plane, as shown
in Figure 17.3. If the direction of motion of this mass is
down the plane, then the frictional force F will act up the
plane, as shown in Figure 17.3, where F = fimg cos 0.
Figure 17.3
Now the weight of the mass is mg and this will cause
two other forces to act on the mass, namely N , and
the component of the weight down the plane, namely
mg sin 0 , as shown by the vector diagram of Figure 17.4.
It should be noted that N acts normal to the surface.
Resolving forces parallel to the plane gives:
Forces up the plane = forces down the plane
i.e. F = mg sin 0 (17.1)
Resolving force perpendicular to the plane gives:
Forces ‘up’ = forces ‘down’
i.e. N = mg cos 0 (17.2)
Dividing equation (17.1) by (17.2) gives:
F mg sin 0 sin#
— = — -=-= tan#
N mg cos # cos #
F
But — = F , hence, tan 0 = p
where g = the coefficient of friction, and 6 = the angle
of repose.
If # is gradually increased until the body starts mo¬
tion down the plane, then this value of # is called the
limiting angle of repose. A laboratory experiment
based on the theory is a useful method of obtaining the
maximum value of g for static friction.
17.5 Motion up a plane with the pulling
force P parallel to the plane
In this case the frictional force F acts down the plane,
opposite to the direction of motion of the body, as
shown in Figure 17.5.
Figure 17.5
The components of the weight mg will be the same as
that shown in Figure 17.4.
Resolving forces parallel to the plane gives:
P = mgsm0 + F (17.3)
Resolving forces perpendicular to the plane gives:
N = mg cos 0 (12.4)
For limiting friction,
F = juN
Figure 17.4 Components of mg
(17.5)
Part Three
Part Three
216 Mechanical Engineering Principles
From equations (17.3) to (17.5), solutions of problems
in this category that involve limiting friction can be
solved.
Determine the value of the force P,
which will just move the body of mass of 25 kg up
the plane shown in Figure 17.6. It may be assumed
that the coefficient of limiting friction, g = 0.3 and
g = 9.81 m/s 2 .
Figure 17.6
From equation (17.4), TV = mg cos # = 25x9.81x cos 15°
= 245.3 x 0.966 = 236.9 N
From equation (17.5), F = juN= 0.3 x 236.9
= 71.1 N
From equation (17.3), P = mg sin 0 + F
= 25 x 9.81 x sin 15°+ 71.1
= 63.48 + 71.1
i.e. force, P = 134.6 N (17.6)
17.6 Motion down a plane with the
pulling force P parallel to the
plane
In this case, the frictional force F acts up the plane,
opposite to the direction of motion of the plane, as
shown in Figure 17.7.
Figure 17.7
The components of the weight mg are shown in Fig¬
ure 17.4, where it can be seen that the normal reaction,
N = mg cos 0, and the component of weight parallel to
and down the plane = mg sin 6
Resolving forces perpendicular to the plane gives:
N = mg cos 0 (17.7)
Resolving forces parallel to the plane gives:
P + mg sin 6 = F (17.8)
When the friction is limiting,
F = juN (17.9)
From equations (17.7) to (17.9), problems arising in
this category can be solved.
If the mass of Problem 7 were
subjected to the force P, which acts parallel to
and down the plane, as shown in Figure 17.7,
determine the value of P to just move the body.
From equation (17.7),
N=mg cos 0 = 25 x 9.81 cos 15°= 236.9 N
From equation (17.9), F = juN= 0.3 x 236.9 = 71.1 N
From equation (17.8), P + mg sin 0 = F
i.e. P + 25 x 9.81 sin 15° - 71.1
i.e. P + 63.5 = 71.1
from which, force,
P= 71.1 -63.5 = 7.6 N (17.10)
From equations (17.6) and (17.10), it can be seen that
the force required to move a body down the plane is so
much smaller than to move the body up the plane.
17.7 Motion up a plane due to a
horizontal force P
This motion, together with the primary forces, is shown
in Figure 17.8.
Figure 17.8
Friction 217
Figure 17.9
In this, the components of mg are as shown in Figure
17.4, and the components of the horizontal force P are
shown by the vector diagram of Figure 17.9.
Resolving perpendicular to the plane gives:
Forces ‘up’ = forces ‘down’
i.e. N = mg cos 0 + P sin 0 (17.11)
Resolving parallel to the plane gives:
P cos 0 = F + mg sin 0 (17.12)
and F = fiN (17.13)
From equations (17.11) to (17.13), problems arising in
this category can be solved.
If the mass of Problem 7 were
subjected to a horizontal force P, as shown in
Figure 17.8, determine the value of P that will
just cause motion up the plane.
Substituting equation (17.13) into equation (17.12)
gives:
or
i.e.
P cos 0 = juN + mg sin 0
juN = P cos 0 - mg sin 0
PcosO mg sin 0
N =
F
F
(17.14)
Equating equation (17.11) and equation (17.14) gives:
P cos 0 mg sin 0
mg cos 0 + P sin 0 =-
F F
i.e. 25 x 9.81 cos 15° + P sin 15°
Pcos 15° 25 x 9.81 sin 15°
“ 03 03
Px 0.966 245.3x 0.259
245.3 x 0.966+Px 0.259 =-—-—-
i.e. 237 + 0.259 P = 3.22 P- 211.8
237 + 211.8 = 3.22 P- 0.259 P
from which, 448.8=2.961 P
448.8
and force P= _ _ - = 151.6 N
2.961
If the mass of Problem 9 were
subjected to a horizontal force P, acting down
the plane, as shown in Figure 17.10, determine
the value of P which will just cause motion
down the plane.
Figure 17.10
The components for mg are shown by the phasor
diagram of Figure 17.4, and the components for P are
shown by the vector diagram of Figure 17.11.
Figure 17.11
Resolving forces down the plane gives:
P cos 0+ mg sin 0 = F (17.15)
Resolving forces perpendicular to the plane gives:
Forces up = forces down
N + P sin 0 = mg cos 6 (17.16)
and F = juN (17.17)
Substituting equation (17.17) into equation (17.15)
gives:
P cos 0 + mg sin 0 = juN
PcosO mg sin 0
from which, N =-1-
F F
From equation (17.16), TV = mg cos 0 - P sin 0
Equating equations (17.18) and (17.19) gives:
(17.18)
(17.19)
PcosO mg sin 0
-1-- = mg cos 0 - P sin 6
F F
Part Three
Part Three
218 Mechanical Engineering Principles
Pcos 15 25 x 9.81sinl5
i.e. -+-
0.3 0.3
-25x9.81 cos 15° -P sin 15°
3.22 P + 211.6 -236.9- 0.259 P
P{ 3.22 + 0.259) = 236.9 - 211.6
3.479 P = 25.3
25.3
from which, force P = ^ ^ = 7.27 N
If in Problem 9, the contact surfaces
were greased, so that the value of [i decreased and
P = 50 N, determine the value of [i which will just
cause motion down the plane.
The primary forces for this problem are shown in
Figure 17.12, where it can be seen that Pis opposite to
the direction of motion.
Figure 17.12
Resolving forces perpendicular to the plane gives:
Forces ‘up’ - forces ‘down’
N = mg cos 0 + P sin 0 (17.20)
Resolving forces parallel to the plane gives:
mg sin 0 - F + P cos 6 (17.21)
and F = juN (17.22)
Substituting equation (17.22) into equation (17.21)
gives:
mg sin 6 = juN + P cos 6 (17.23)
Substituting equation (17.20) into equation (17.23)
gives:
mg sin 0 - (i{mg cos 0 + P sin 0) + P cos 0
Now try the following Practice Exercise
Practice Exercise 96 Further problems on
friction on an inclined
plane
(Where necessary, take g = 9.81 m/s 2 )
1. A mass of 40 kg rests on a flat horizontal
surface as shown in Figure 17.13. If the
coefficient of friction g = 0.2, determine
the minimum value of a horizontal force
P which will just cause it to move.
[78.48 N]
Motion
-►
^ r>
* H
>
rr
ig
Figure 17.13
2. If the mass of Problem 1 were equal to
50 kg, what will be the value of PI
[98.1 N]
3. An experiment is required to obtain the
static value of //; this is achieved by in¬
creasing the value of 6 until the mass just
moves down the plane, as shown in Figure
17.14. If the experimentally obtained value
for 0 was 22.5°, what is the value of //?
[ju = 0.414]
Figure 17.14
i.e. 25x9.81 sin 15° =//(25 x 9.81 cos 15° +
50 sin 15°)+ 50 cos 15°
Hence 63.48 = ju( 236.89 + 12.94) + 48.3
63.48 - 48.3 =jux 249.83
from which,
15.18
/l = 249 Y 3 = °-° 61
4. If in Problem 3, g was 0.6, what would be
the experimental value of 01 [0 = 30.96°]
5. For a mass of 50 kg just moving up an
inclined plane, as shown in Figure 17.5,
what would be the value of P, given that
0 - 20° and n = 0.4? [P - 352.1 N]
Friction 219
6. For a mass of 50 kg, just moving down an
inclined plane, as shown in Figure 17.7,
what would be the value of P, given that
0 = 20° and p = 0.4? [P = 16.6 N]
7. If in Problem 5, 6 = 10° and p = 0.5, what
would be the value of PI [P = 326.7 N]
8. If in Problem 6, 0 — 10° and p = 0.5, what
would be the value of PI [P = 156.3 N]
9. Determine P for Problem 5, if it were
acting in the direction shown in Figure
17.8. [P = 438.6 N]
10. Determine P for Problem 6, if it were act¬
ing in the direction shown in Figure 17.10.
[P= 15.43 N]
11. Determine the value for 6 which will
just cause motion down the plane, when
P = 250 N and acts in the direction shown
in Figure 17.12. It should be noted that
in this problem, motion is down the plane,
p = 0.5 and m = 50 kg.
[6= 53.58°]
12. If in Problem 11, <9 = 30°, determine the
value of [i. [p = 0.052]
17.8 The efficiency of a screw jack
Screw jacks (see Section 20.4, page 243) are often used
to lift weights; one of their most common uses are to
raise cars, so that their wheels can be changed. The the¬
ory described in Section 17.7 can be used to analyse
screw jacks.
Consider the thread of the square-threaded screw
j ack shown in Figure 17.15.
Let p be the pitch of the thread, i.e. the axial distance
that the weight W is lifted or lowered when the screw is
turned through one complete revolution.
From Figure 17.15, the motion of the screw in lifting
the weight can be regarded as pulling the weight by a
horizontal force P, up an incline 0 , where
P
tan 0= —: as shown in Figure 17.15,
nd
and
(A + d 2 )
2
If p is the coefficient of friction up the slope, then let
tan A = p
Referring now to Figure 17.16, the screw jack can be
analysed.
Figure 17.16
Resolving normal to the plane gives:
N= W cos 0 + P sin 6 (17.24)
Resolving parallel to the plane gives:
P cos 0 = F+ W sin 6 (17.25)
and F = pN (17.26)
Substituting equation (17.26) into equation (17.25)
gives:
P cos 6 = pN + W sin 6 (17.27)
D z
< ►
Substituting equation (17.24) into equation (17.27)
gives:
P cos 0 = p(W cos 6 + P sin 0) + W sin 6
Dividing each term by cos 6 and remembering that
sin 0
-— = tan 0 gives:
cos 6
P = p(W+P tan 6) + Wtan 0
Rearranging gives:
Figure 17.15
P( 1 -p tan 0)=W(p + tan 0)
Part Three
Part Three
220 Mechanical Engineering Principles
from which, P =
W(p + tan#) W(tan A + tan #)
(1 - p tan #) (1 - tan A tan #)
since /1 = tan A
However, from compound angle formulae (see refer¬
ence [1] on page 222),
(tan A + tan #)
tan(2 + #)
(1 - tan A tan #)
Hence,
P=W tan (# + 2)
(17.28)
However, from Figure 17.15,
P
tan # = —7 and tan 2 = u
nd
(
hence P =
IF (tan A + tan#)
1 - tan A tan #
W
P +
P
nd
\
/
2
1 -
v
/r<i
\
/
(17.29)
Multiplying top and bottom of equation (17.29) by
nd gives:
W(pnd + p)
P =
(nd - pp)
(17.30)
The useful work done in lifting the weight W a
distance of p = Wp (17.31)
From Figure 17.15, the actual work done = Pxnd
W(pnd + p)
(nd - pp)
x nd
(17.32)
• useful work done u • ,,
Einciency rj = - which is usually
actual work done
expressed as a percentage
Wp
i.e.
p(nd - pp)
^ Wjpnd + p) x nd (pnd + p)x nd
(nd- pp)
Dividing throughout by nd gives:
(
P
1-
ri =
v
pp_
nd
\
/
p( 1 - tan A tan 6)
(pnd + p)
/
nd
P +
v
P
nd
\
However, tan(2 + 6)
p( 1 - tan A tan 0)
/r<i(tanA + tan#)
tan A + tan 6
from com-
(1 - tan A tan 6)
pound angle formulae (see reference [1], page 222)
Hence,
but
n
P 1
nd tan(A + 0)
P
nd
= tan 6
hence, efficiency, i/
tan#
tan(A + #)
(17.33)
From equations (17.31) and (17.32),
the work lost in friction
W(pnd + p)
(nd - pp)
nd - Wp
(17.34)
The coefficient of friction on the
sliding surface of a screw jack is 0.2. If the pitch
equals 1 cm, and IJ ] =4 cm and Z) ? = 5 cm, calcu¬
late the efficiency of the screw jack.
Working in millimetres, d
P
tan #
_ (D { + D 2 ) _ (40 + 50)
2 2
= 45 mm,
1 cm =10 mm,
p 10
nd n x 45
= 0.0707,
from which, # = tan -1 (0.0707) = 4.05°
and tan 2 = p = 0.2,
from which, 2 = tan -1 (0.2) = 11.31°
From equation (17.33),
tan#
efficiency
n
tan(A + #)
0.0707
0.0707
i.e.
tan(l 1.31 + 4.05)° 0.2747
= 0.257
rj = 25.7%
Now try the following Practice Exercises
Practice Exercise 97
Further problem on
the efficiency of a
screw jack
1. The coefficient of friction on the sliding sur¬
face of a screw jack whose thread is similar
to Figure 17.15, is 0.24. If the pitch equals
12 mm, and = 42 mm and D 2 = 56 mm,
calculate the efficiency of the screw jack.
[24.06%]
Friction 221
Practice Exercise 98 Short-answer
questions on friction
1. The.of frictional force depends on the
.of surfaces in contact.
2. The.of frictional force depends on the
size of the.to the surfaces in contact.
3. The . of frictional force is always
.to the direction of motion.
4. The coefficient of friction between surfaces
should be a.value for materials con¬
cerned with bearings.
5. The coefficient of friction should have a
. value for materials concerned with
braking systems.
6. The coefficient of dynamic or sliding fric¬
tion is given by —
7. The coefficient of static or limiting friction
is given by — when.is just about to
take place.
8. Lubricating surfaces in contact result in a
.of the coefficient of friction.
9. Briefly discuss the factors affecting the size
and direction of frictional forces.
10. Name three practical applications where a
low value of coefficient of friction is desir¬
able and state briefly how this is achieved
in each case.
11. Name three practical applications where
a high value of coefficient of friction is
required when transmitting forces and
discuss how this is achieved.
12. For an object on a surface, two differ¬
ent values of coefficient of friction are
possible. Give the names of these two
coefficients of friction and state how their
values may be obtained.
13. State the formula for the angle of repose.
14. What theory can be used for calculating the
efficiency of a screw jack.
Practice Exercise 99 Multiple-choice
questions on friction
(Answers on page 336)
1. A block of metal requires a frictional force
F to keep it moving with constant velocity
across a surface. If the coefficient of friction
is then the normal force N is given by:
(a) j (b) fiF
(c) L (d) F
N
2. The unit of the linear coefficient of friction
is:
(a) newtons
(b) radians
(c) dimensionless
(d) newtons/metre
Questions 3 to 7 refer to the statements given
below. Select the statement required from each
group given.
(a) The coefficient of friction depends on the
type of surfaces in contact.
(b) The coefficient of friction depends on the
force acting at right angles to the surfaces
in contact.
(c) The coefficient of friction depends on the
area of the surfaces in contact.
(d) Frictional force acts in the opposite direc¬
tion to the direction of motion.
(e) Frictional force acts in the direction of
motion.
(f) A low value of coefficient of friction is
required between the belt and the wheel in
a belt drive system.
(g) A low value of coefficient of friction is
required for the materials of a bearing.
(h) The dynamic coefficient of friction is
given by (normal force)/(frictional force)
at constant speed.
(i) The coefficient of static friction is given
by (applied force)/(frictional force) as slid¬
ing is just about to start.
(j) Lubrication results in a reduction in the
coefficient of friction.
3. Which statement is false from (a), (b), (f)
and (!)?
Part Three
Part Three
222 Mechanical Engineering Principles
4. Which statement is false from (b), (e), (g)
and (j)?
5. Which statement is true from (c), (f), (h)
and (i)?
6. Which statement is false from (b), (c), (e)
and (j)?
7. Which statement is false from (a), (d), (g)
and (h)?
8. The normal force between two surfaces
is 100 N and the dynamic coefficient of
friction is 0.4
The force required to maintain a constant
speed of sliding is:
(a) 100.4 N (b) 40 N
(c) 99.6 N (d) 250 N
9. The normal force between two surfaces is
50 N and the force required to maintain
a constant speed of sliding is 25 N. The
dynamic coefficient of friction is:
(a) 25 (b) 2
(c) 75 (d) 0.5
10. The maximum force, which can be applied
to an object without sliding occurring, is
60 N, and the static coefficient of friction
is 0.3. The normal force between the two
surfaces is:
(a) 200 N (b) 18 N
(c) 60.3 N (d) 59.7 N
11. The formula for the angle of repose is:
(a) F = juN (b) tan 9 = ju
F sin 0
(c) // = — (d) tan 0
N
cos 6
Reference
[1] BIRD, J. O. Engineering Mathematics 7th Edition , chap¬
ter 27, Taylor and Francis Publishers, 2014.
For fully worked solutions to each of the problems in Practice Exercises 95 to 99 in this chapter,
go to the website:
www.routledge.com/cw/bird
Chapter 18
Motion in a circ e
Why it is important to understand: Motion in a circle
In this chapter, uniform circular motion of particles is considered, and it is assumed that objects such as
railway trains and motorcars behave as particles. When a railway train goes round a bend, its wheels
will have to produce a centripetal acceleration towards the centre of the turning circle. This in turn will
cause the railway tracks to experience a centrifugal thrust, which will tend to cause the track to move
outwards. To avoid this unwanted outward thrust on the outer rail, it will be necessary to incline the
railway tracks. Although problems involving the motion in a circle are dynamic ones, they can be reduced
to static problems through D’Alembert’s principle. If a motorcar travels around a bend, its tyres will have
to exert centripetal forces to achieve this; this is achieved by the transverse frictional forces acting on the
tyres. Problems involving locomotives and cars travelling around bends, a conical pendulum and motion in
a vertical circle are solved in this chapter, and the centrifugal clutch is explained. This chapter is therefore
of importance in the study of the behaviour of structures and bodies undergoing circular motion.
At the end of this chapter you should be able to:
• understand centripetal force
• understand D’Alembert’s principle
• understand centrifugal force
• solve problems involving locomotives and cars travelling around bends
• solve problems involving a conical pendulum
• solve problems involving the motion in a vertical circle
• understand the centrifugal clutch
18.1 Introduction
In this chapter we will restrict ourselves to the uni¬
form circular motion of particles. We will assume that
objects such as railway trains and motorcars behave as
particles, i.e. rigid body motion is neglected. When a
railway train goes round a bend, its wheels will have
to produce a centripetal acceleration towards the centre
of the turning circle. This in turn will cause the railway
tracks to experience a centrifugal thrust (see below),
which will tend to cause the track to move outwards.
To avoid this unwanted outward thrust on the outer rail,
it will be necessary to incline the railway tracks in the
manner shown in Figure 18.1.
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Three
224 Mechanical Engineering Principles
From Section 15.3, it can be seen that when a particle
moves in a circular path at a constant speed v, its cen¬
tripetal acceleration,
0
a = 2 v sin —
x
1
t
When 6 is small, 6 ~ sin 0 ,
hence
However,
Therefore
6 1 6
a- 2v - x - = v -
2 t t
e
co = uniform angular velocity = —
a = v co
If r = the radius of the turning circle, then
v = co r
v 2
and a = co 2 r = —
r
In Figure 18.2, the following notation is used:
CG = centre of gravity of
the car,
m = mass of car,
R x = vertical reaction of
‘inner’ wheel,
F x = frictional force on
‘inner’ wheel,
h = vertical distance
of the centre of
gravity of the car
from the ground,
L = distance between
the centre of the
tyres,
// = coefficient of
friction,
CF = centrifugal force
_ mv 2
5
r
g = acceleration due to
gravity,
R 7 = vertical reaction of
‘outer’ wheel,
F 2 = frictional force on
‘outer’ wheel,
r = radius of the
turning circle.
Now force = mass x acceleration
2
Ttl V **
Hence, centripetal force = m co 2 r = —-— (18.1)
D’Alembert’s principle
Although problems involving the motion in a circle are
dynamic ones, they can be reduced to static problems
through D’Alembert’s principle. In this principle,
the centripetal force is replaced by an imaginary
centrifugal force which acts equal and opposite to the
centripetal force. By using this principle, the dynamic
problem is reduced to a static one.
If a motorcar travels around a bend, its tyres will
have to exert centripetal forces to achieve this. This is
achieved by the transverse frictional forces acting on
the tyres, as shown in Figure 18.2.
Figure 18.2
Determine expressions for the
frictional forces F { and F 2 of Figure 18.2. Hence
determine the thrust on each tyre.
mv
Resolving forces horizontally gives:
F x + F 2 = centrifugal force = CF
~ i
Resolving forces vertically gives:
R ] + R 2 = mg
Taking moments about the ‘outer’ wheel gives:
CF x h + R j x L = mg —
(18.2)
(18.3)
i.e.
mv 2 7 _ _ L
— h+ R { L = mg —
or
Hence,
from which,
Also,
L mv 2
R ] L = mg— - h
z r
( 9 )
gL v z h
R\L = m
v
2
/
/
n - m
r ~l
\
gL v 2 h
v
/
(18.4)
F x =/iR x and F 2 = jdR 1 (18.5)
Substituting equation (18.4) into equation (18.3) gives:
/
m
T
gL v 2 h
v
/
+ R 2 = mg
/
Therefore,
m
R 2 = mg- j
gL v 2 h '
V
/
Motion in a circle 225
mg m v 2 h
-f + T~
i.e.
D — m
r i~T
f o ^
gL v 2 h
— +-
(18.6)
v
From equations (18.4) to (18.6):
/ - \
m
gL v 2 h
V
/
and
F,= fi
m
\
r
gL v 2 h
— + -
2 r
v
To calculate the thrust on each tyre:
From Pythagoras’ theorem (see Chapter 1),
(18.7)
(18.8)
T|= v F i 2 + R \ = 4 ^ 2r \ + R \
i.e.
T x =R { x\l + M 2 (see Figure 18.3(a))
(a)
Figure 18.3
Let a x = angle of thrust,
(b)
/ \
i.e.
aj= tan
-l
F,
l
— fn n 1
tan ‘//
From Figure 18.3(b), T 2 = + R i
= V) l 2 R 1 +
= X J1 + /A
a 2 = tan
-l
/ \
h.
r 7
v 2 7
— 4-n n 1
tan A //
18.2 Motion on a curved banked track
A railway train is required to travel
around a bend of radius r at a uniform speed of v.
Determine the amount that the ‘outer’ rail is to be
elevated to avoid an outward centrifugal thrust in
these rails, as shown in Figure 18.4.
\
\
\
\
\
\
To balance the centrifugal force:
(t?! + R 1 ) sin 6 = CF =
mv
from which,
Let R = R X +
sin 6 =
mv
r(R\ + R 2 )
Then
sin 6 =
mv
~rR
(18.9)
Resolving forces vertically gives:
R cos 0 = mg
from which,
R =
mg
cos 0
(18.10)
Substituting equation (18.10) into equation (18.9)
gives:
mv*-
sin 6= -cos 0
Hence
tan 6 =
r mg
v
rg
, . sin#
(since-= tan 0 )
cos 0
Thus, the amount that the outer rail has to be elevated
to avoid an outward centrifugal thrust on these
rails,
6 = tan
-l
(18.11)
A locomotive travels around a curve
of 700 m radius. If the horizontal thrust on the outer
rail is l/40th of the locomotive’s weight, determine
the speed of the locomotive (in km/h). The surface
that the rails are on may be assumed to be horizon¬
tal and the horizontal force on the inner rail may be
assumed to be zero. Take g as 9.81 m/s 2 .
Part Three
Part Three
226 Mechanical Engineering Principles
mg
Centrifugal force on outer rail =
mg
Hence,
from which,
mv
v
40
2- 8 r -
9.81x700
40 40
= 171.675 m 2 /s 2
i.e. v = a/171.675 = 13.10 m/s
= (13.10 x 3.6) km/h
i.e. the speed of the locomotive, v = 47.17 km/h
What angle of banking of the rails is
required for Problem 3 above, for the outer rail to
have a zero value of thrust? Assume the speed of
the locomotive is 40 km/h.
From Problem 2, angle of banking, 6= tan
v = 40 km/h
km lh 1000m
= 40 — x-x
-l
f 2^
40
T6
h 3600s 1km
= 11.11 m/s
Hence, 6 = tan
-l
11.11 2 m 2 /s 2
v
700m x 9.81 m/s 2
— tan — 1
tan _1 (0.01798)
i.e. angle of banking, 6 = 1.03*
Now try the following Practice Exercise
Practice Exercise 100 Further problems on
motion in a circle
(Where needed, take g = 9.81 m/s 2 )
1. A locomotive travels around a curve of
500 m radius. If the horizontal thrust on the
outer rail is of the locomotive weight,
w/ V/
determine the speed of the locomotive. The
surface that the rails are on may be assumed
to be horizontal and the horizontal force on
the inner rail may be assumed to be zero.
[35.64 km/h]
2. If the horizontal thrust on the outer rail
of Problem 1 is of the locomotive’s
weight, determine its speed.
[25.2 km/h]
3. What angle of banking of the rails of Prob¬
lem 1 is required for the outer rail to have
a zero value of outward thrust? Assume the
speed of the locomotive is 15 km/h.
[0.203°]
4. What angle of banking of the rails is required
for Problem 3, if the speed of the locomotive
is 30 km/h? [0.811°]
18.3 Conical pendulum
If a mass m were rotated at a constant angular velocity,
co , in a horizontal circle of radius r, by a mass-less taut
string of length L , its motion will be in the form of a
cone, as shown in Figure 18.5.
Figure 18.5 Conical pendulum
Let r = radius of horizontal turning circle,
L = length of string,
h = OC,
co = constant angular velocity about C,
m = mass of particle P,
T— tension in string,
and 6 = cone angle
Determine an expression for the
cone angle 6 and the tension in the string T, for the
conical pendulum of Figure 18.5. Determine also an
expression for co.
Motion in a circle 227
Resolving forces horizontally gives:
Centrifugal force = CF = T sin 0
m co 2 r = T sin 6
i.e.
from which,
T =
mco 2 r
sin 0
Resolving forces vertically gives:
T cos 0 = mg
mg
(18.12)
from which,
T =
cos 0
(18.13)
Equating equations (18.12) and (18.13) gives:
Rearranging gives:
i.e.
From Figure 18.5,
mco 2 r
mg
sin 6
cos 0
mco 2 r
sin 0
mg
cos 0
co 2 r
tan 6 =
g
e, 0 =
tan -1
sin 0 =
n (18.12),
T=
/ 2 ^
co z r
g
\
r
L
mco 2 r
(18.14)
(18.15)
r
L
i.e. the tension in the string, T = m co 2 L (18.16)
From equation (18.14),
But, from Figure 18.5,
Hence,
and
co 2 r
g
= tan 6
tan 6= —
n
co 2 r
g
r
h
co 2 = f
h
/ g
Thus, angular velocity about C,co= J—
n
(18.17)
A conical pendulum rotates about a
horizontal circle at 90 rpm. If the speed of rotation
of the mass increases by 10%, how much does
the mass of the pendulum rise (in mm)? Take g
as 9.81 m/s 2 .
Angular velocity,
Inn In x 90
co
60 60
= 9.425 rad/s
I g
From equation (18.17), co = J—
h
or
® 2 = f
h
from which, height, h
g
9.81
co 2 9.425 2
= 0.11044 m (see Figure 18.5)
When the speed of rotation rises by 10%,
n 2 = 90 x 1.1 = 99 rpm
Hence, co 7
2nn 2 _ In x 99
60
60
= 10.367 rad/s
From equation (18.17), co 2
g
h 2
or
co
2 _ g
h 2 =
h 2
g
9.81
Hence, , 0
2 col 10 ' 367
i.e. the new value of height, h 2 = 0.09127 m
Rise in height of the pendulum mass = ‘old’ h - ‘new’ h
= /z — /z 2 = 0.11044 — 0.09127
= 0.01917 m = 19.17 mm
A conical pendulum rotates at a hori¬
zontal angular velocity of 5 rad/s. If the length of
the string is 2 m and the pendulum mass is 0.3 kg,
determine the tension in the string. Determine also
the radius of the turning circle. Take g as 9.81 m/s 2 .
Angular velocity, co = 5 rad/s
From equation (18.16), tension in the string,
T = m co 2 L
= 0.3 kg x (5 rad/s) 2 x 2 m
i.e. T= 15 kg m/s 2
However, 1 kg m/s 2 = 1 N, hence,
tension in the string, T= 15 N
mg
From equation (18.13), T
from which,
cos 6
cos 9
mg
T
0.3kg x 9.81 m/s
15N
= 0.1962
Hence, the cone angle, 6= cos _1 (0.1962)
= 78.685°
r
From equation (18.15), sin 0 = —
from which, radius of turning circle,
r = L sin 6 = 2 m x sin 78.685°
= 1.961 m
Part Three
Part Three
228 Mechanical Engineering Principles
Now try the following Practice Exercise
Practice Exercise 101 Further problems on
the conical pendulum
1. A conical pendulum rotates about a
horizontal circle at 100 rpm. If the speed of
rotation of the mass increases by 5%, how
much does the mass of the pendulum rise?
[8.32 mm]
2. If the speed of rotation of the mass of Prob¬
lem 1 decreases by 5%, how much does the
mass fall? [9.66 mm]
3. A conical pendulum rotates at a horizontal
angular velocity of 2 rad/s. If the length of
the string is 3 m and the pendulum mass is
0.25 kg, determine the tension in the string.
Determine also the radius of the turning
circle. [3 N, 1.728 m]
18.4 Motion in a vertical circle
This problem is best solved by energy considerations.
Consider a particle P rotating in a vertical circle
of radius r, about a point O , as shown in Figure 18.6.
Neglect losses due to friction.
Figure 18.6 Motion in vertical circle
Let T = tension in a mass-less string,
r = radius of turning circle,
and m = mass of particle.
Determine the minimum tangential
velocity at A, namely, v A , which will just keep the
string taut at the point B for the particle moving in
the vertical circle of Figure 18.6.
At the point B the potential energy = mg x 2 r (18.18)
2
and the kinetic energy = - (18.19)
2
At the point A, the kinetic energy (. KE) = — A - (18.20)
and the potential energy ( PE) = 0
As there are no energy losses, KE at A = (KE + PE) at B
Hence, from equations (18.18) to (18.20):
or
mv A
mg x 2 r +
mv B
+ 2 gr
from which, v a = V B + ^S r
(18.21)
At B, T= 0; this is because B is the highest point in the
circle, where it can readily be observed that T will be
a minimum
Thus,
oi-
weight = centrifugal force at B ,
mv B 2
mg= -
from which, v B 2 = gr
(18.22)
Substituting equation (18.22) into equation (18.21)
gives:
V A 2 = g>' + 4gr = 5 gr
Hence, the minimum tangential velocity at^4,
v A = Eir ( 18 . 23 )
A mass of 0.1 kg is being rotated
in a vertical circle of radius 0.6 m. If the mass is
attached to a mass-less string and the motion is
such that the string is just taut when the mass is at
the top of the circle, what is the tension in the string
when it is horizontal? Neglect losses and take g
as 9.81 m/s 2 .
Motion in a circle 229
At the top of the circle, potential energy
PE = 2 mgr
2
and
where
KE =
mvj
v T = velocity at the top
When the string is horizontal,
PE = mgr
2
mv.
and kinetic energy, KE
z
where Vj = velocity of mass at this point.
From the conservation of energy,
{PE + KE) at the top = {PE + KE) when the string is
horizontal
i.e.
or
i.e.
2 mgr = mgr +
mv\
mvj
mv i 2 mvr 2
—-— = 2 mgr - mgr +
2
Vi
but CF at top
gr +
mvj
Vj'
mg or v T 2 = gr
Hence,
v f
gr +
gr _ 3 gr
i.e.
vi
2 _
3 gr
and vj = -\[3gr = V3 x 9.81 x 0.6 = 4.202 m/s
Resolving forces horizontally,
Centrifugal force = T = tension in the string
mvf 0.1kg x (4.202) 2 m 2 /s 2
Therefore,
T=
l
0.6m
i.e. the tension in the string, T = 2.943 N
What is the tension in the string for
Problem 9 when the mass is at the bottom of the
circle?
From equation (18.23), the velocity at the bottom of the
circle = v = ->j5gr
i.e. v= \l 5 x 9.81 x 0.6 = 5.4249 m/s.
Resolving forces vertically,
T = tension in the string
= centrifugal force + the weight of the mass
/
i.e. T =
mv
+ mg = m
v
+ g
= 0.1 x
V y
5 4249 2
———- + 9.81
0.6
v y
= 0.1 x (49.05 + 9.81)
= 0.1 x 58.86 N
i.e. tension in the string, T= 5.886 N
If the mass of Problem 9 were to
rise, so that the string is at 45° to the vertical axis
and below the halfway mark, what would be the
tension in the string?
At 45°, PE = '-gf-
and
KE =
mv 2
where v 2 = velocity of the mass at this stage.
From the conservation of energy, {PE + KE) at the top
= {PE + KE) at this stage
Therefore,
2mg r= m ^ + m ^-
mvj
~ir
From Problem 9, v T 2 = gr
2
or
v 2 '
r \
~ r r
2 r -1—
2 2
v z A J
g
2gr
from which,
and
V 2 2 = 4 gr
Vr
= Fgr
= + x 9.81x0.6
= 4.852 m/s
Resolving forces in a direction along the string,
T = tension in the string = centrifugal
force + component of weight at 45° to
the vertical
i.e.
T =
mv 2
+ mg cos 45°
0.1 x (4.2852)
06
+ 0.1 x 9.81 x 0.7071
= 3.924 N + 0.6937 N
i.e. the tension in the string, T = 4.618 N
Part Three
Part Three
230 Mechanical Engineering Principles
Now try the following Practice Exercise
Practice Exercise 102 Further problems on
motion in a vertical
circle
1. A uniform disc of diameter 0.1 m rotates
about a vertical plane at 200 rpm. The disc
has a mass of 1.5 kg attached at a point on
its rim and another mass of 2.5 kg at another
point on its rim, where the angle between the
two masses is 90° clockwise. Determine the
magnitude of the resultant centrifugal force
that acts on the axis of the disc, and its posi¬
tion with respect to the 1.5 kg mass.
[63.94 N at 59° clockwise]
2. If a mass of 4 kg is placed on some posi¬
tion on the disc in Problem 1, determine the
position where this mass must be placed to
nullify the unbalanced centrifugal force.
At a radius of 36.44 mm,
121° anti-clockwise to 1.5 kg mass
3. A stone of mass 0.1 kg is whirled in a verti¬
cal circle of 1 m radius by a mass-less string,
so that the string just remains taut. Deter¬
mine the velocity and tension in the string at
(a) the top of the circle (b) the bottom of the
circle (c) midway between (a) and (b).
(a) 3.132 m/s, ON (b) 7 m/s, 5.88 N'
(c) 5.42 m/s, 2.94 N
18.5 Centrifugal clutch
A clutch is an engineering device used for transferring
motion from an engine to a gearbox or other machin¬
ery. The main purpose of the clutch is to transfer the
motion in a smooth and orderly manner, so that the
gears and wheels (in the case of the motor car) will
accelerate smoothly and not in a jerky manner.
The centrifugal clutch works on the principle that
the rotating driving shaft will cause the centrifugal
weights, shown in Figure 18.7, to move radially out¬
wards with increasing speed of rotation of the driving
shaft. These centrifugal weights will be restrained by
the restraining springs shown, but when the speed of
the driving shaft reaches the required value, the clutch
material will engage with the driven shaft, through fric¬
tion, and cause the driven shaft to rotate. The driven
Figure 18.7
shaft will thus reach a high speed of rotation quite
smoothly in the required time.
Centrifugal clutches are popular when it is required to
exert a high starting torque quickly and smoothly.
A suitable clutch material is asbestos, but it is likely
that asbestos will be replaced by more modern materi¬
als for health and safety reasons.
Now try the following Practice Exercises
Practice Exercise 103 Short-answer
problems on motion
in a circle
1. The centrifugal force of a mass m moving at
velocity v at a radius r is given by:.
2. What is the potential energy at the top of a
circle for the motion in a vertical circle?
3. What is the potential energy at the bottom of
a circle for the motion in a vertical circle?
4. What is the potential energy at the ‘middle’
of a circle for the motion in a vertical circle?
Exercise 104 Multiple-choice problems on
motion in a circle
(Answers on page 336)
1. To decrease the horizontal thrust on the outer
rail of a train going round a bend, the outer
rail should be:
(a) lowered
(b) raised
(c) kept at the same level as the inner rail
(d) made bigger
Motion in a circle 231
2. If the speed of rotation of a conical pendu¬
lum is increased, the height of the pendulum
mass will:
(a) fall
(b) become zero
(c) stay the same
(d) rise
3. The minimum tension on the top of a vertical
circle, for satisfactory motion in a circle is:
(a) zero (b) mg
2
m v
(c) - (d) negative
r
4. If v is the velocity at the ‘middle’ for the
motion in a vertical circle, the tension is:
mv 2
(a) zero (b) -
r
(c) mg (d) negative
5. If the tension in the string is zero at the
top of a circle for the motion in a vertical
circle, the velocity at the bottom of the
circle is:
(a) zero (b) y[5gr
(c) yfgr (d) yj3gr
For fully worked solutions to each of the problems in Practice Exercises 100 to 104 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Three
Chapter 19
Why it is important to understand: Simple harmonic motion
Simple harmonic motion is of importance in a number of branches of engineering and physics, includ¬
ing structural and machine vibrations, alternating electrical currents, sound waves, light waves, tidal
motion, and so on. This chapter explains simple harmonic motion, determines natural frequencies for
spring-mass systems, calculates periodic times and explains simple and compound pendulums. SHM
is of particular importance in electrical engineering and in the vibration and periodic oscillations of
structures and bodies.
At the end of this chapter you should be able to:
• understand simple harmonic motion
• determine natural frequencies for simple spring-mass systems
• calculate periodic times
• understand the motion of a simple pendulum
• understand the motion of a compound pendulum
19.1 Introduction to simple harmonic
motion (SHM)
A particle is said to be under SHM if its acceleration
along a line is directly proportional to its displacement
along that line, from a fixed point on that line.
Consider the motion of a particle A, rotating in a
circle with a constant angular velocity co , as shown in
Figure 19.1.
Consider now the vertical displacement of A from xx,
as shown by the distance y c . If P is rotating at a con¬
stant angular velocity co then the periodic time T to
travel an angular distance of 2/r, is given by:
2k
T=— (19.1)
co
Let /= frequency of motion C (in hertz), where
Displacement (y c )
1 CO
f- T = 2* (19 ' 2)
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
To determine whether or not SHM is taking place, we
will consider motion of A in the direction yy. Now
y c = OA sin cot,
Simple harmonic motion 233
i.e. y c = r sin cot (19.3)
where t = time in seconds.
Plotting of equation (19.3) against t results in
the sinusoidal variation for displacement, as shown in
Figure 19.1(b).
From Chapter 13, v 4 = cor , which is the tangential ve¬
locity of the particle A.
From the velocity vector diagram, at the point A on the
circle of Figure 19.1(a),
v c = v A c ° s 0= v 4 c ° s cot (19.4)
Plotting of equation (19.4) against t results in the
sinusoidal variation for the velocity v c , as shown in
Figure 19.1(b).
The centripetal acceleration of A
= a 4 = Qrr
Now a c = -a A smO
Therefore, a c = -co 1 r sin cot (19.5)
Plotting of equation (19.5) against t results in the sinu¬
soidal variation for the acceleration at C, a c , as shown
in Figure 19.1 (b).
Substituting equation (19.3) into equation (19.5) gives:
a c = -co 2 y c (19.6)
Equation (19.6) shows that the acceleration along
the line yy is directly proportional to the displace¬
ment along this line, therefore the point C is moving
with SHM.
2 ji
Now T = —, but from equation (19.6),
co
9 2 a
a r =co A y , i.e. co - —
c y
Therefore,
i.e.
T=27i
dispacement
acceleration
In general, from equation (19.6),
a + co 2 y = 0
(19.7)
19.2 The spring-mass system
(a) Vibrating horizontally
Consider a mass m resting on a smooth surface and at¬
tached to a spring of stiffness k , as shown in Figure 19.2.
Figure 19.2
If the mass is given a small displacement x, the spring
will exert a resisting force of kx, i.e. F = — foe
But, F = ma
hence, ma = -kx
or ma + kx = 0
or
k
a + —x = 0
m
(19.8)
Equation (19.8) shows that this mass is oscillating (or vi¬
brating) in SHM, or according to equation (19.7). Com¬
paring equation (19.7) with equation (19.8) we see that
k
from which,
Now
T=
2k
m
-2 nA—
co V k
and/= frequency of oscillation or vibration
f - m 1
i.e.
2k
(19.9)
(b) Vibrating vertically
Consider a mass m , supported by a vertical spring
of stiffness k , as shown in Figure 19.3. In this equi¬
librium position, the mass has an initial downward
static deflection ofy 0 . If the mass is given an additional
downward displacement of y and then released, it will
vibrate vertically.
Figure 19.3
Part Three
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234 Mechanical Engineering Principles
The force exerted by the spring = -k(y Q +y)
Therefore, F = accelerating force - resisting force
= ma
or F= mg - k(y o + y) = ma
i.e. F = mg - ky o ~ ky = ma
But, ky o = mg ,
hence F = mg - mg - ky = ma
Thus, ma + ky = 0
or
a+ —y = 0
m
i.e. SHM takes place, then co = J — and T
m
i.e. periodic time,
and frequency,
t = 2 7 rjj
f-
co
1
2k 2k
as before (from equation (19.9)).
k
m
2k
CO
(19.10)
(19.11)
Comparing equations (19.9) and (19.11), it can be seen
that there is no difference in whether the spring is hori¬
zontal or vertical.
A mass of 1.5 kg is attached to a
vertical spring, as shown in Figure 19.4. When the
mass is displaced downwards a distance of 55 mm
from its position of rest, it is observed to oscillate
60 times in 72 seconds. Determine (a) periodic
time (b) the stiffness of the spring (c) the time taken
to travel upwards a distance of 25 mm for the first
time (d) the velocity at this point.
Figure 19.4
Periodic time, T =
72 seconds
60 oscillations
= 1.2 seconds
From equation (19.10),
i.e.
T=2ttJ-
1.2 = 2 Ka
F5
k
Hence,
from which,
1.2 2 = ( 2 tt) 2 x
k = (2 n) 1 x
L5
k
1.5
1 . 2 2
i.e. stiffness of spring, k = 41.1 N/m
(r ~ 25)
(c) From Figure 19.4, cos 0= -
r
(55 - 25)
or cos 0 = -—-= 0.545
from which, 0= cos _1 0.545 =56.94°
2k 2k
Now, co =
= 5.236 rad/s
T 1.2
But 0 = cot , hence, time t taken to travel upwards
a distance of 25 mm, is given by:
6
56.94° 2k rad
t
x
CO rad 360°
5.236-
= 0.19 s
(d) Velocity at C in Figure 19.4,
v c = v 4 sin 0
= cor sin 6
ra d 55 .
= 5.236 — x-m x sm56.94°
s 1000
= 0.288 x 0.838 m/s
i.e. v c = 0.241 m/s after 25 mm of travel
Now try the following Practice Exercise
Practice Exercise 105 Further problems
on simple harmonic
motion
1. A particle oscillates 50 times in 22 s. Deter¬
mine the frequency and periodic time.
|/=2.27 Hz, T=0.44 s]
2. A yacht floats at a depth of 2.2 m. On a par¬
ticular day, at a time of 09.30 h, the depth at
low tide is 1.8 m and at a time of 17.30 h, the
depth of water at high tide is 3.4 m. Deter¬
mine the earliest time of day that the yacht
is refloated. [12 h, 10 min, 1 s]
3. A mass of 2 kg is attached to a vertical
spring. The initial state displacement of this
mass is 74 mm. The mass is displaced down¬
wards and then released. Determine (a) the
stiffness of the spring, and (b) the frequency
of oscillation of the mass.
[(a) 265.1 N/m (b) 1.83 Hz]
Simple harmonic motion 235
4. A particle of mass 4 kg rests on a smooth
horizontal surface and is attached to a hori¬
zontal spring. The mass is then displaced
horizontally outwards from the spring a dis¬
tance of 26 mm and then released to vibrate.
If the periodic time is 0.75 s, determine
(a) the frequency / (b) the force required to
give the mass the displacement of 26 mm,
(c) the time taken to move horizontally
inwards for the first 12 mm.
[(a) 1.33 Hz (b) 7.30 N (c) 0.12 s]
5. A mass of 3 kg rests on a smooth horizontal
surface, as shown in Figure 19.5. If the stiff¬
ness of each spring is 1 kN/m, determine the
frequency of vibration of the mass. It may
be assumed that, initially, the springs are
un-stretched.
From Section 15.4, page 193,
T= I o a = -restoring couple
= -mg(L sin 6)
But, I () = mL 2 = mass moment of inertia about the point
of rotation
hence, mL? a + mgL sin 6=0
For small deflections, sin 6=0
Hence, l 2 a + gL6=0
or
But
Therefore,
a+ cc?6= 0 (from equation 19.7)
g
L
/
AAA/WWW
/
-A/WVWWV
Figure 19.5
[4.11 Hz]
6. A helical spring has a mass of 10 kg attached
to its top. If the mass vibrates vertically
with a frequency of 1.5 Hz, determine the
stiffness of the spring. [888.3 N/m]
19.3 The simple pendulum
A simple pendulum consists of a particle of mass m
attached to a mass-less string of length T, as shown in
Figure 19.6.
and
Now
(19.12)
(19.13)
and
(19.14)
If the simple pendulum of Figure
19.6 were of length 2 m, determine its frequency of
vibration. Take g = 9.81 m/s 2 .
g
From equation (19.14), frequency,/= — 2 -
2 n
= 0.352 Hz
In order to determine the value of g
at a certain point on the Earth’s surface, a simple
pendulum is used. If the pendulum is of length 3 m
and its frequency of oscillation is 0.2875 Hz,
determine the value of g.
From equation (19.14), frequency, f=
Figure 19.6
i.e.
g
0.2875 = 11
2 n
Part Three
Part Three
236 Mechanical Engineering Principles
and (0.2875) 2 x(2tt) 2 =|
3.263 = |
from which, acceleration due to gravity, g = 3 x 3.263
= 9.789 m/s
19.4 The compound pendulum
Consider the compound pendulum of Figure 19.7,
which oscillates about the point O. The point G in
Figure 19.7 is the position of the pendulum’s centre of
gravity.
y
Figure 19.7
Let I () = mass moment of inertia about O
Now T = I o cc = - restoring couple
= -mgh sin 6
From the parallel axis theorem,
I n = I - mh 2 = mk r 2
Go G
or I = mk r 2 + mh 2
O G
where I G = mass moment of inertia about G,
k G 2 = radius of gyration about G
Now T= I o a
or - mgh sin 6 = I o a
but I 0 = mk G 2 + mh 2
Therefore, [mk G -\- mh 2 ^a =-mgh sin 0
but for small displacements, sin 6=6
Hence,
+ h 2 )a = -mgh 6
i.e.
or
k G + /? 2 )a +gh6= 0
a +
gh
k 2 + h 2
0 =0
However,
a+ co 2 0= 0
but this motion is simple harmonic motion (see equa¬
tion (19.7))
? gh
Therefore, co z = ~r~i -vT
ik d +h-)
and
and
T- — = in JM±Z)
CO y gh
_1 _ 1 1 gh
f = T~ 2n\ (k 2 G + h 2 )
(19.15)
(19.16)
(19.17)
It is required to determine the mass
moment of inertia about G of a metal ring, which
has a complex cross-sectional area. To achieve
this, the metal ring is oscillated about a knife edge,
as shown in Figure 19.8, where the frequency of
oscillation was found to be 1.26 Hz. If the mass
of the ring is 10.5 kg, determine the mass moment
of inertia about the centre of gravity, I G . Take
g= 9.81 m/s 2 .
y
Figure 19.8
Simple harmonic motion 237
By inspection of Figure 19.8, h = 15 mm = 0.075 m.
1
Now frequency, f z
gh
2n^(kl + h 2 )
i.e.
1.26 =
1 9.81x0.075
27r]l(k2 + 0.075 2 )
i.e.
/, „\ 2 1 9.81x0.075
(1.26) =--x
from which, + 0.005625 -
(2 n) z \k G + 0.075 2 j
0.73575
1.5876 x (2 n) 2
= 0.011739
k G 2 = 0.011739 -0.005625
= 0.006114
from which, /t G = V0.006114 =0.0782
The mass moment of inertia about the centre of gravity,
I G = m k 2 = 10.5 kg x 0.006114 m 2
i.e. I G = 0.0642 kg m 2
19.5 Torsional vibrations
From equation (19.7), it can be seen that for SHM in a
linear direction,
a + co 2 y = 0
For SHM in a rotational direction,
ar + co 2 y = 0
( \
or
a+ cq 2
y_
= 0
V )
or
a+ co 2 0 = 0
(see equation (19.7))
i.e
• •
9 + co 2 9~ 0
(19.18)
2. What will be the period of oscillation if
g = 9.78 m/s 2 for the pendulum of Problem 1?
[0.3519 Hz]
3. What will be the period of oscillation if
g = 9.832 m/s 2 for the pendulum of Problem 1 ?
[0.3529 Hz]
4. What will be the value of the mass moment
of inertia through the centre of gravity, I G ,
for the compound pendulum of worked
problem 4 on page 236, if the inner diameter
of the disc of Figure 19.8 were 100 mm?
[0.0559 kg m 2 ]
Practice Exercise 107 Short-answer
questions on simple
harmonic motion
1. State the relationship between the displace¬
ment (y) of a mass and its acceleration
(a) for SHM to take place.
2. State the relationship between frequency /
and periodic time T when SHM takes place.
3. State the formula for the frequency of oscil¬
lation for a simple pendulum.
4. State a simple method of increasing the
period of oscillation of the pendulum of a
‘grandfather’ clock.
Practice Exercise 108 Multiple-choice
questions on simple
harmonic motion
(Answers on page 336)
y
where 0 = — = angular displacement, and
r
0 = a = angular acceleration
Now try the following Practice Exercises
Practice Exercise 106 Further problems on
pendulums
1. Determine the period of oscillation of a
pendulum of length 2 m if g = 9.81 m/s 2 .
[0.3525 Hz]
1. Tidal motion is normally related to which
mathematical function?
(a) tangent (b) sine
(c) square root (d) straight line
2. If the mass of a simple pendulum is doubled,
its period of oscillation:
(a) increases (b) decreases
(c) stays the same (d) doubles
3. A pendulum has a certain frequency of
oscillation in London. Assuming that tem¬
perature remains the same, the frequency of
Part Three
Part Three
238 Mechanical Engineering Principles
oscillation of the pendulum if it is measured
on the equator:
(a) increases
(b) decreases
(c) remains the same
(d) doubles
4. The period of oscillation of a simple pendu¬
lum of length 9.81 m, given g = 9.81 m/s 2 is:
(a) 6.28 Hz
(b) 0.455 Hz
(c) 17.96 Hz
(d) 0.056 Hz
For fully worked solutions to each of the problems in Practice Exercises 105 to 108 in this chapter,
go to the website:
www.routledge.com/cw/bird
Chapter 20
Why it is important to understand: Simple machines
This chapter commences by defining load, effort, mechanical advantage, velocity ratio and efficiency,
where efficiency is then defined in terms of mechanical advantage and velocity ratio. These terms, together
with other terms defined in earlier chapters, are applied to simple and quite complex pulley systems,
screw-jacks, gear trains and levers. This chapter is fundamental to the study of the behaviour of machines
and is thus important in studying the fundamental concepts of machine motion and their behaviour.
At the end of this chapter you should be able to:
• define a simple machine
• define force ratio, movement ratio, efficiency and limiting efficiency
• understand and perform calculations with pulley systems
• understand and perform calculations with a simple screw-jack
• understand and perform calculations with gear trains
• understand and perform calculations with levers
20.1 Machines
A machine is a device that can change the magnitude
or line of action, or both magnitude and line of action
of a force. A simple machine usually amplifies an input
force, called the effort, to give a larger output force,
called the load. Some typical examples of simple
machines include pulley systems, screw-jacks, gear
systems and lever systems.
20.2 Force ratio, movement
ratio and efficiency
The force ratio or mechanical advantage is defined
as the ratio of load to effort, i.e.
Force ratio = ^, aC * = mechanical advantage (20. 1)
effort
Since both load and effort are measured in newtons,
force ratio is a ratio of the same units and thus is a
dimension-less quantity.
The movement ratio or velocity ratio is defined as the
ratio of the distance moved by the effort to the distance
moved by the load, i.e.
m „ . distance moved by the effort
Movement ratio = ——-——-—-——
distance moved by the load
= velocity ratio (20.2)
Since the numerator and denominator are both mea¬
sured in metres, movement ratio is a ratio of the same
units and thus is a dimension-less quantity.
The efficiency of a simple machine is defined as the
ratio of the force ratio to the movement ratio, i.e.
„ _ . force ratio
Efficiency =-—
movement ratio
mechanical advantage
velocity ratio
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Three
240 Mechanical Engineering Principles
Since the numerator and denominator are both
dimension-less quantities, efficiency is a dimension-less
quantity. It is usually expressed as a percentage, thus:
Efficiency =
force ratio
movement ratio
x 100 %
(20.3)
Due to the effects of friction and inertia associated
with the movement of any object, some of the input
energy to a machine is converted into heat and losses
occur. Since losses occur, the energy output of a ma¬
chine is less than the energy input, thus the mechanical
efficiency of any machine cannot reach 100%.
For simple machines, the relationship between effort
and load is of the form: F e = aF l + Z?, where F e is the
effort, F/ is the load and a and b are constants. From
equation (20.1),
load _ Fj _ F l
force ratio = _ - — - — ",
effort F e aF/ + b
Dividing both numerator and denominator by F/ gives:
F, _ 1
aF, + b b
1 a H-
f,
When the load is large, F, is large and — is small com-
b/
pared with a. The force ratio then becomes approximate¬
ly equal to ^ and is called the limiting force ratio, i.e.
limiting ratio = ^
The limiting efficiency of a simple machine is defined
as the ratio of the limiting force ratio to the movement
ratio, i.e.
Limiting efficiency
a x movement ratio
x 100 %
From equation (20.2),
_ distance moved by the effort
movement ratio distance moved by the load
16 m
1.6 m
From equation (20.3),
* forCe rati0 w , AAO/
efficiency = -— x 100%
movement ratio
7.84
10
= 78.4%
x 100
For the simple machine of Problem
1, determine: (a) the distance moved by the effort
to move the load through a distance of 0.9 m
(b) the effort which would be required to raise a
load of 200 kg, assuming the same efficiency
(c) the efficiency if, due to lubrication, the effort
to raise the 160 kg load is reduced to 180 N.
(a) Since the movement ratio is 10, then from equa¬
tion (20.2),
distance moved by the effort
= 10 x distance moved by the load
= 10 x 0.9 = 9 m
Since the force ratio is 7.84, then from equation
( 20 . 1 ),
effort =
load
T84
200x9.8
7.84
= 250 N
(c) The new force ratio is given by
where a is the constant for the law of the machine:
F e = aF, + b
Due to friction and inertia, the limiting efficiency of
simple machines is usually well below 100%.
A simple machine raises a load of
160 kg through a distance of 1.6 m. The effort applied
to the machine is 200 N and moves through a distance
of 16 m. Taking g as 9.8 m/s 2 , determine the force
ratio, movement ratio and efficiency of the machine.
From equation (20.1),
load
force ratio = ——
effort
160kg
200N
160x9.81N
200N
= 7.84
j^l = M^=8.7n
effort 1^0
Hence, the new efficiency after lubrication
8.711
= —— x 100 = 87.11%
In a test on a simple machine, the
effort/load graph was a straight line of the form
F e = aF x + b. Two values lying on the graph
were at F e = 10 N, Fj = 30 N, and at F e = 74 N,
Fj = 350 N. The movement ratio of the machine
was 17. Determine: (a) the limiting force ratio
(b) the limiting efficiency of the machine.
(a) The equation F e = aF/ + b is of the formy = mx + c,
where m is the gradient of the graph. The slope
Simple machines 241
of the line passing through points (xj, y x ) and
(x 2 , y 2 ) of the graph y = mx + c is given by:
yi-y\
m
x 2 -x {
Thus for F e = aF { + b, the slope a is given by:
74-10
a
350 -30
64 = 0.2
320
The limiting force ratio is — = = 5
a 0.2
(b) The limiting efficiency
1
a x movement ratio
1
x 100
0.2 x 17
x 100 =29.4%
Now try the following Practice Exercise
Practice Exercise 109 Further problems on
force ratio, movement
ratio and efficiency
1. A simple machine raises a load of 825 N
through a distance of 0.3 m. The effort is
250 N and moves through a distance of
3.3 m. Determine: (a) the force ratio (b) the
movement ratio (c) the efficiency of the ma¬
chine at this load. [(a) 3.3 (b) 11 (c) 30%]
2. The efficiency of a simple machine is 50%.
If a load of 1.2 kN is raised by an effort of
300 N, determine the movement ratio. [8]
3. An effort of 10 N applied to a simple ma¬
chine moves a load of 40 N through a dis¬
tance of 100 mm, the efficiency at this load
being 80%. Calculate: (a) the movement
ratio (b) the distance moved by the effort.
[(a) 5 (b) 500 mm]
4. The effort required to raise a load using a
simple machine, for various values of load is
as shown:
(b) the limiting force ratio (c) the limiting
efficiency.
[(a) F e = 0.04 F x + 170 (b) 25 (c) 83.3%]
5. For the data given in question 4, determine
the values of force ratio and efficiency for
each value of the load. Hence plot graphs
of effort, force ratio and efficiency to a base
of load. From the graphs, determine the
effort required to raise a load of 6 kN and
the efficiency at this load. [410 N, 48%]
20.3 Pulleys
A pulley system is a simple machine. A single-pulley
system, shown in Figure 20.1(a), changes the line
of action of the effort, but does not change the mag¬
nitude of the force. A two-pulley system, shown in
Figure 20.1(b), changes both the line of action and the
magnitude of the force.
If the movement ratio for the machine is
30, determine (a) the law of the machine
Figure 20.1
(b)
Part Three
Part Three
242 Mechanical Engineering Principles
Figure 20.1 (Continued)
Theoretically, each of the ropes marked (i) and (ii) share
the load equally, thus the theoretical effort is only half
of the load, i.e. the theoretical force ratio is 2. In prac¬
tice the actual force ratio is less than 2 due to losses.
A three-pulley system is shown in Figure 20.1(c). Each
of the ropes marked (i), (ii) and (iii) carry one-third of
the load, thus the theoretical force ratio is 3. In general,
for a multiple pulley system having a total of n pulleys,
the theoretical force ratio is n. Since the theoretical
efficiency of a pulley system (neglecting losses) is
100 and since from equation (20.3),
efficiency =
force ratio x 1Q0%
movement ratio
it follows that when the force ratio is n,
100 =
-- x 100%
movement ratio
that is, the movement ratio is also n.
A load of 80 kg is lifted by a three-
pulley system similar to that shown in Figure
20.1(c) and the applied effort is 392 N. Calculate
(a) the force ratio (b) the movement ratio (c) the
efficiency of the system. Take g to be 9.8 m/s 2 .
(a) From equation (20.1), the force ratio is given by
load
effort
(b) From above, for a system having n pulleys, the
movement ratio is n. Thus, for a three-pulley sys¬
tem, the movement ratio is 3
From equation (20.3),
efficiency =
force ratio
movement ratio
x 100%
2
= -x 100-66.67%
A pulley system consists of
two blocks, each containing three pulleys and
connected as shown in Figure 20.2. An effort
of 400 N is required to raise a load of 1500 N.
Determine (a) the force ratio (b) the movement
ratio (c) the efficiency of the pulley system.
Figure 20
(a) From equation (20.1),
(b)
(c)
force ratio =
load 1500
effort 400
= 3.75
An ^-pulley system has a movement ratio of n ,
hence this 6-pulley system has a movement ratio
of 6
From equation (20.3),
force ratio
efficiency = -— x 100%
movement ratio
3.75
= x 100 = 62.5%
Now try the following Practice Exercise
Practice Exercise 110 Further problems on
pulleys
The load is 80 kg, i.e. (80 x 9.8) N, hence, 1. A pulley system consists of four pulleys in
„ . 80 x 9.8 an upper block and three pulleys in a lower
force ratio = ——— =2
y a *
Simple machines 243
block. Make a sketch of this arrangement
showing how a movement ratio of 7 may be
obtained. If the force ratio is 4.2, what is the
efficiency of the pulley? [60%]
2. A three-pulley lifting system is used to raise a
load of 4.5 kN. Determine the effort required
to raise this load when losses are neglected. If
the actual effort required is 1.6 kN, determine
the efficiency of the pulley system at this load.
[1.5 kN, 93.75%]
20.4 The screw-jack
A simple screw-jack is shown in Figure 20.3 and is a
simple machine since it changes both the magnitude
and the line of action of a force (see also Section 17.8).
Load
The screw of the table of the jack is located in a fixed
nut in the body of the jack. As the table is rotated by
means of a bar, it raises or lowers a load placed on the
table. For a single-start thread, as shown, for one com¬
plete revolution of the table, the effort moves through
a distance 2nr , and the load moves through a distance
equal to the lead of the screw, say, L.
Movement ratio =
2nr
~L
(20.4)
For the efficiency of a screw-jack, see Section 17.8,
page 219.
A screw-jack is being used to support
the axle of a car, the load on it being 2.4 kN. The
screw jack has an effort of effective radius 200 mm
and a single-start square thread, having a lead of
5 mm. Determine the efficiency of the jack if an
effort of 60 N is required to raise the car axle.
From equation (20.3),
efficiency
force ratio
movement ratio
x 100%
load 2400 N
where force ratio = ——— = ——— = 40
effort
60 N
From equation (20.4),
2k r 2^(200) mm
movement ratio
Hence, efficiency
L
5 mm
= 251.3
force ratio
movement ratio
x 100%
40
251.3
x 100 = 15.9%
Now try the following Practice Exercise
Practice Exercise 111 Further problems on
the screw-jack
1. Sketch a simple screw-jack. The single-start
screw of such a jack has a lead of 6 mm and
the effective length of the operating bar from
the centre of the screw is 300 mm. Calcu¬
late the load which can be raised by an effort
of 150 N if the efficiency at this load is 20%.
[9.425 kN]
2. A load of 1.7 kN is lifted by a screw-jack
having a single-start screw of lead 5 mm.
The effort is applied at the end of an arm of
effective length 320 mm from the centre of
the screw. Calculate the effort required if the
efficiency at this load is 25%. [16.91 N]
20.5 Gear trains
A simple gear train is used to transmit rotary motion
and can change both the magnitude and the line of
action of a force, hence it is a simple machine. The gear
train shown in Figure 20.4 consists of spur gears and
has an effort applied to one gear, called the driver, and
a load applied to the other gear, called the follower.
In such a system, the teeth on the wheels are so
spaced that they exactly fill the circumference with a
whole number of identical teeth, and the teeth on the
driver and follower mesh without interference. Under
these conditions, the number of teeth on the driver and
Part Three
Part Three
244 Mechanical Engineering Principles
.9
wi/
C 0 )
7 .
%
Driver
oJV Vj
):«?
Follower
Figure 20.4
follower are in direct proportion to the circumference
of these wheels, i.e.
number of teeth on driver
number of teeth on follower
circumference of driver
circumference of follower (20-5)
If there are, say, 40 teeth on the driver and 20 teeth on
the follower then the follower makes two revolutions
for each revolution of the driver. In general:
number of revolutions made by driver
number of revolutions made by the follower
number of teeth on follower ^ ^
-:- ( 20 . 6 )
number of teeth on driver
It follows from equation (20.6) that the speeds of the
wheels in a gear train are inversely proportional to the
number of teeth. The ratio of the speed of the driver
wheel to that of the follower is the movement ratio, i.e.
speed of driver
Movement ratio - speed of follower “
teeth on follower ^
teeth on driver
When the same direction of rotation is required on both
the driver and the follower an idler wheel is used as
shown in Figure 20.5.
Driver (A) Idler ( B ) Follower (C)
Figure 20.5
Let the driver, idler, and follower be A, B and C,
respectively, and let Abe the speed of rotation and T be
the number of teeth. Then from equation (20.7),
Eb
*A
T
A
T
or N A - N B
T
B
B
T
A
and
Thus,
N
c
N
T r T r
B or N c = N B B
B
T.
c
T
c
speed of A
speed of C N
, _ T r
n b b
T
A
C
t b
x
Ar T r
1 c
T C -T C
T a T
B
T
A
This shows that the movement ratio is independent
of the idler, only the direction of the follower is being
altered.
A compound gear train is shown in Figure 20.6, in
which gear wheels B and C are fixed to the same shaft
and hence N B = N c
Figure 20.6
N
From equation (20.7), —A - JL i.e. N B - N A x
T
A
Also,
But N b = N c and
N b T a
n d _t c
T
B
N c T d
i.e. N d =N c x T c
' D
T
N d = N b x Zl
1 D
therefore,
T T
N D = N A x^ x _f. ( 20 . 8 )
A
T B t d
For compound gear trains having, say, P gear wheels,
T A Tr
Tr
Tr
N P = N A x — x x _T~ .x °
A
t b T d T f
T
p
from which,
. N A
movement ratio = jy
Simple machines 245
A driver gear on a shaft of a motor
has 35 teeth and meshes with a follower having
98 teeth. If the speed of the motor is 1400 revolu¬
tions per minute, find the speed of rotation of the
follower.
From equation (20.7),
speed of driver _ teeth on follower
speed of follower teeth on driver
1400 98
i c - = —
speed of follower 35
1400x35
Hence, speed of follower =-—- = 500 rev/min
98
A compound gear train similar to
that shown in Figure 20.6 consists of a driver gear
A, having 40 teeth, engaging with gear B , having
160 teeth. Attached to the same shaft as B , gear C
has 48 teeth and meshes with gear D on the output
shaft, having 96 teeth. Determine (a) the move¬
ment ratio of this gear system and (b) the efficiency
when the force ratio is 6.
From equation (20.8), the speed
T a T.
- speed of A x — x
T B t
From equation (20.7), movement ratio
_ speed of A _ Tg Tp_ _ 160 96
speed of D T A T c 40 X 48
The efficiency of any simple machine
force ratio
movement ratio
x 100%
Thus,
efficiency = - x 100 = 75%
Now try the following Practice Exercise
2. A compound gear train has a 30-tooth driver
gear A , meshing with a 90-tooth follower
gear B. Mounted on the same shaft as B and
attached to it is a gear C with 60 teeth, mesh¬
ing with a gear D on the output shaft having
120 teeth. Calculate the movement and force
ratios if the overall efficiency of the gears is
72%. [6,4.32]
3. A compound gear train is as shown in Figure
20.6. The movement ratio is 6 and the num¬
bers of teeth on gears A, C and D are 25, 100
and 60, respectively. Determine the number
of teeth on gear B and the force ratio when
the efficiency is 60%. [250,3.6]
20.6 Levers
A lever can alter both the magnitude and the line of ac¬
tion of a force and is thus classed as a simple machine.
There are three types or orders of levers, as shown in
Figure 20.7.
Load
F ,
Effort
Effort
Load Fe
F i
(b) Fulcrum
Practice Exercise 112 Further problems on
gear trains
1. The driver gear of a gear system has 28 teeth
and meshes with a follower gear having 168
teeth. Determine the movement ratio and the
speed of the follower when the driver gear
rotates at 60 revolutions per second.
[6, 10 rev/s]
Effort
F e Load
(c) Fulcrum
Figure 20.7
Part Three
Part Three
246 Mechanical Engineering Principles
A lever of the first order has the fulcrum placed
between the effort and the load, as shown in Figure
20.7(a).
A lever of the second order has the load placed be¬
tween the effort and the fulcrum, as shown in Figure
20.7(b).
A lever of the third order has the effort applied
between the load and the fulcrum, as shown in Figure
20.7(c).
Problems on levers can largely be solved by apply¬
ing the principle of moments (see Chapter 6). Thus for
the lever shown in Figure 20.7(a), when the lever is in
equilibrium,
anticlockwise moment = clockwise moment
i.e. a x F x = b x F e
Thus, force ratio =
b
a
distance of effort from fulcrum
distance of load from fulcrum
The load on a first-order lever,
similar to that shown in Figure 20.7(a), is 1.2 kN.
Determine the effort, the force ratio and the move¬
ment ratio when the distance between the fulcrum
and the load is 0.5 m and the distance between the
fulcrum and effort is 1.5 m. Assume the lever is
100% efficient.
Applying the principle of moments, for equilibrium:
anticlockwise moment = clockwise moment
i.e. 1200 N x 0.5 m = effort x 1.5 m
1200x0.5
Hence, effort =-—- = 400 N
„ . Fj 1200 „
force ratio . . _ . 3
Alternatively, force ratio = — = —- = 3
a 0.5
This result shows that to lift a load of, say, 300 N, an
effort of 100 N is required.
Since, from equation (20.3),
■ force ratio 1AAn/
efficiency = -— x 100%
movement ratio
t . force ratio 1AAO/
then, movement ratio = __ . -x l uu%
efficiency
3
This result shows that to raise the load by, say, 100 mm,
the effort has to move 300 mm.
A second-order lever, AB , is in a
horizontal position. The fulcrum is at point C. An
effort of 60 N applied at B just moves a load at
point D, when BD is 0.75 m and BC is 1.25 m.
Calculate the load and the force ratio of the lever.
A second-order lever system is shown in Figure
20.7(b). Taking moments about the fulcrum as the load
is just moving, gives:
anticlockwise moment = clockwise moment
i.e. 60 N x 1.25 m = load x 0.75 m
Thus,
60x1.25
load = . - = 100 N
0.75
From equation (20.1),
, load
force ratio =
100
effort 60
= 1.67
Alternatively,
force ratio = d^ance °f effort from fulcrum
distance of load from fulcrum
1.25
0.75
= 1.67
Now try the following Practice Exercises
Practice Exercise 113 Further problems
on levers
1. In a second-order lever system, the force
ratio is 2.5. If the load is at a distance of
0.5 m from the fulcrum, find the distance
that the effort acts from the fulcrum if
losses are negligible. [1.25 m]
2. A lever AB is 2 m long and the fulcrum is at
a point 0.5 m from B. Find the effort to be
applied at A to raise a load of 0.75 kN at B
when losses are negligible. [250 N]
3. The load on a third-order lever system is at a
distance of 750 mm from the fulcrum and the
effort required to just move the load is 1 kN
when applied at a distance of 250 mm from
the fulcrum. Determine the value of the load
and the force ratio if losses are negligible.
[333.3 N, 1/3]
100
x 100 = 3
Simple machines 247
Practice Exercise 114 Short-answer
questions on simple
machines
1. State what is meant by a simple machine.
2. Define force ratio.
3. Define movement ratio.
4. Define the efficiency of a simple machine
in terms of the force and movement ratios.
5. State briefly why the efficiency of a simple
machine cannot reach 100%.
6. With reference to the law of a simple ma¬
chine, state briefly what is meant by the
term ‘limiting force ratio’.
7. Define limiting efficiency.
8. Explain why a four-pulley system has a
force ratio of 4, when losses are ignored.
9. Give the movement ratio for a screw-jack
in terms of the effective radius of the effort
and the screw lead.
10. Explain the action of an idler gear.
11. Define the movement ratio for a two-gear
system in terms of the teeth on the wheels.
12. Show that the action of an idler wheel
does not affect the movement ratio of a
gear system.
13. State the relationship between the speed of
the first gear and the speed of the last gear
in a compound train of four gears, in terms
of the teeth on the wheels.
14. Define the force ratio of a first-order lever
system in terms of the distances of the load
and effort from the fulcrum.
15. Use sketches to show what is meant by:
(a) a first-order (b) a second-order (c) a
third-order lever system. Give one practi¬
cal use for each type of lever.
Practice Exercise 115 Multiple-choice
questions on simple
machines
(Answers on page 336)
A simple machine requires an effort of 250 N
moving through 10 m to raise a load of 1000 N
through 2 m. Use this data to find the correct an¬
swers to questions 1 to 3, selecting these answers
from:
(a) 0.25 (b) 4
(c) 80% (d) 20%
(e) 100 (f) 5
(g) 100% (h) 0.2
(i) 25%
1. Find the force ratio.
2. Find the movement ratio.
3. Find the efficiency.
4.
5.
6 .
7.
The law of a machine is of the form
F e = aF x + b. An effort of 12 N is required to
raise a load of 40 N and an effort of 6 N is
required to raise a load of 16 N. The move¬
ment ratio of the machine is 5. Use this data
to find the correct answers to questions 4 to
6, selecting these answers from:
(a) 80% (b) 4
(c) 2.8 (d) 0.25
(e) F (f) 25%
(g) 100% (h) 2
(i) 25%
Determine the constant ‘a\
Find the limiting force ratio.
Find the limiting efficiency.
Which of the following statements is false?
(a) A single-pulley system changes the
line of action of the force but does not
change the magnitude of the force,
when losses are neglected.
(b) In a two-pulley system, the force ratio
is ]- when losses are neglected.
(c) In a two-pulley system, the movement
ratio is 2.
(d) The efficiency of a two-pulley system
is 100% when losses are neglected.
8. Which of the following statements con¬
cerning a screw-jack is false?
(a) A screw-jack changes both the line of
action and the magnitude of the force.
(b) For a single-start thread, the distance
moved in 5 revolutions of the table is
5f, where £ is the lead of the screw.
Part Three
Part Three
248 Mechanical Engineering Principles
(c) The distance moved by the effort is
2n r, where r is the effective radius of
the effort.
(d) The movement ratio is given by
2 nr
~TF
9. In a simple gear train, a follower has
50 teeth and the driver has 30 teeth. The
movement ratio is:
(a) 0.6 (b) 20
(c) 1.67 (d) 80
10. Which of the following statements is
true?
(a) An idler wheel between a driver and a
follower is used to make the direction
of the follower opposite to that of the
driver.
(b) An idler wheel is used to change the
movement ratio.
(c) An idler wheel is used to change the
force ratio.
(d) An idler wheel is used to make the
direction of the follower the same as
that of the driver.
11. Which of the following statements is false?
(a) In a first-order lever, the fulcrum is
between the load and the effort.
(b) In a second-order lever, the load is
between the effort and the fulcrum.
(c) In a third-order lever, the effort is ap¬
plied between the load and the fulcrum.
(d) The force ratio for a first-order lever
system is given by:
distance of load from fulcrum
distance of effort from fulcrum
12. In a second-order lever system, the load
is 200 mm from the fulcrum and the effort
is 500 mm from the fulcrum. If losses are
neglected, an effort of 100 N will raise a
load of:
(a) 100 N (b) 250 N
(c) 400 N (d) 40 N
For fully worked solutions to each of the problems in Practice Exercises 109 to 115 in this chapter,
go to the website:
www.routledge.com/cw/bird
Revision Test 7 Friction, motion in a circle, simple harmonic motion and simple machines
This Revision Test covers the material contained in Chapters 17 to 20. The marks for each question are shown in
brackets at the end of each question.
Assume, where necessary, that the acceleration due to
gravity, g = 9.81 m/s 2
1. The material of a brake is being tested and it is
found that the dynamic coefficient of friction
between the material and steel is 0.90. Calcu¬
late the normal force when the frictional force is
0.630 kN. (4)
2. A mass of 10 kg rests on a plane, which is in¬
clined at 30° to the horizontal. The coefficient of
friction between the mass and the plane is 0.6.
Determine the magnitude of a force, applied par¬
allel to and up the plane, which will just move the
mass up the plane. (10)
3. If in Problem 2, the force required to just move
the mass up the plane, is applied horizontally,
what will be the minimum value of this force?
( 10 )
4. A train travels around a curve of radius 400 m.
If the horizontal thrust on the outer rail is to be
l/30th the weight of the train, what is the velocity
of the train (in km/h)? It may be assumed that the
inner and outer rail rails are on the same level and
that the inner rail takes no horizontal thrust. (6)
5. A conical pendulum of length 2.5 m rotates in a
horizontal circle of diameter 0.6 m. Determine its
angular velocity. (6)
6. Determine the time of oscillation for a simple
pendulum of length 1.5 m. (6)
7. A simple machine raises a load of 120 kg through
a distance of 1.2 m. The effort applied to the ma¬
chine is 150 N and moves through a distance of
12 m. Taking g as 10 m/s 2 , determine the force
ratio, movement ratio and efficiency of the
machine. (6)
8. A load of 30 kg is lifted by a three-pulley system
and the applied effort is 140 N. Calculate, taking
g to be 9.8 m/s 2 , (a) the force ratio (b) the move¬
ment ratio (c) the efficiency of the system. (5)
9. A screw-jack is being used to support the axle of a
lorry, the load on it being 5.6 kN. The screw jack
has an effort of effective radius 318.3 mm and a
single-start square thread, having a lead of 5 mm.
Determine the efficiency of the jack if an effort of
70 N is required to raise the car axle. (6)
10. A driver gear on a shaft of a motor has 32 teeth
and meshes with a follower having 96 teeth. If
the speed of the motor is 1410 revolutions per
minute, find the speed of rotation of the follower.
(4)
11. The load on a first-order lever is 1.5 kN. Deter¬
mine the effort, the force ratio and the movement
ratio when the distance between the fulcrum and
the load is 0.4 m and the distance between the
fulcrum and effort is 1.6 m. Assume the lever is
100% efficient. (7)
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 7,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
Part Three
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Part Four
Heat Transfer and
Fluid Mechanics
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Chapter 21
Why it is important to understand: Heat energy and transfer
This chapter defines sensible and latent heat and provides the appropriate formulae to calculate the
amount of energy required to convert a solid to a gas and vice-versa, and also for other combinations
of solids, liquids and gases. This information is often required by engineers if they are required to
design an artefact (say) to convert ice to steam via the state of liquid water. An example of a household
requirement of when this type of calculation is required is that of the simple domestic kettle. When the
designer is required to design a domestic electric kettle, it is important that the design is such that the
powering arrangement is (just) enough to boil the required amount of water in a reasonable time. If the
powering were too low, you may have great difficulty in boiling the water when the kettle is full. Similar
calculations are required for large water containers, which are required to boil large quantities of water
for other uses, including for kitchens in schools, to make tea/coffee, etc., and for large hotels, which
have many uses for hot water. The chapter also describes the three main methods of heat transfer,
namely conduction, convection and radiation, together with their uses. All of this is fundamental in the
design of heat engines, compressors, refrigerators, etc.
At the end of this chapter you should be able to:
• distinguish between heat and temperature
• appreciate that temperature is measured on the Celsius or the thermodynamic scale
• convert temperatures from Celsius into kelvin and vice versa
• recognise several temperature measuring devices
• define specific heat capacity, c and recognise typical values
• calculate the quantity of heat energy Q using Q = mc(t 2 -1 j)
• understand change of state from solid to liquid to gas, and vice versa
• distinguish between sensible and latent heat
• define specific latent heat of fusion
• define specific latent heat of vaporisation
• recognise typical values of latent heats of fusion and vaporisation
• calculate quantity of heat Q using Q = mL
• describe the principle of operation of a simple refrigerator
21.1 Introduction
Heat is a form of energy and is measured in joules.
Temperature is the degree of hotness or coldness of a
substance. Heat and temperature are thus not the same
thing. For example, twice the heat energy is needed to
boil a full container of water than half a container - that
is, different amounts of heat energy are needed to cause
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Four
254 Mechanical Engineering Principles
an equal rise in the temperature of different amounts of
the same substance.
Temperature is measured either (i) on the Celsius*
(°C) scale (formerly Centigrade), where the tem¬
perature at which ice melts, i.e. the freezing point of
water, is taken as 0°C and the point at which water boils
under normal atmospheric pressure is taken as 100°C,
or (ii) on the thermodynamic scale, in which the unit
of temperature is the kelvin* (K). The Kelvin scale
*Anders Celsius (27 November 1701-25 April 1744) was a
Swedish astronomer that went on to propose the Celsius tem¬
perature scale, which takes his name, in 1742. To find out more
go to www.routledge.com/cw/bird
* William Thomson, 1st Baron Kelvin (26 June 1824-17
December 1907) was a mathematical physicist and engineer.
Knighted by Queen Victoria for his work on the transatlantic
telegraph project, Kelvin also had extensive maritime interests
and was most noted for his work on the mariner’s compass. To
find out more go to www.routledge.com/cw/bird
uses the same temperature interval as the Celsius scale
but as its zero takes the ‘absolute zero of temperature’
which is at about - 273°C. Hence,
kelvin temperature = degree Celsius + 273
i.e. K - (°C) + 273
Thus, for example, 0°C = 273 K, 25°C - 298 K and
100°C = 373 K
Convert the following temperatures
into the Kelvin scale:
(a) 37°C (b) - 28°C
From above, kelvin temperature = degree Celsius + 273
(a) 37°C corresponds to a kelvin temperature of
37+ 273, i.e. 310 K
(b) -28°C corresponds to a kelvin temperature of
-28+ 273, i.e. 245 K
Convert the following temperatures
into the Celsius scale:
(a) 365 K (b) 213 K
From above, K - (°C) + 273
Hence, degree Celsius = kelvin temperature -273
(a) 365 K corresponds to 365 - 273, i.e. 92°C
(b) 213 K corresponds to 213 - 273, i.e. - 60°C
Now try the following Practice Exercise
Practice Exercise 116 Further problems on
temperature scales
1. Convert the following temperatures into the
Kelvin scale:
(a) 51°C (b) -78°C
(c) 183°C
[(a) 324 K (b) 195 K (c) 456 K]
2. Convert the following temperatures into the
Celsius scale:
(a) 307 K (b) 237 K
(c) 415 K
[(a) 34°C (b) -36°C (c) 142°C]
21.2 The measurement of
temperature
A thermometer is an instrument that measures
temperature. Any substance that possesses one or more
Heat energy and transfer 255
properties that vary with temperature can be used to
measure temperature. These properties include changes in
length, area or volume, electrical resistance or in colour.
Examples of temperature measuring devices include:
(i) liquid-in-glass thermometer, which uses the
expansion of a liquid with increase in tempera¬
ture as its principle of operation,
(ii) thermocouples, which use the e.m.f. set up,
when the junction of two dissimilar metals is
heated,
(iii) resistance thermometer, which uses the change
in electrical resistance caused by temperature
change, and
(iv) pyrometers, which are devices for measuring
very high temperatures, using the principle that
all substances emit radiant energy when hot, the
rate of emission depending on their temperature.
Each of these temperature measuring devices, together
with others, are described in Chapter 26, page 312.
21.3 Specific heat capacity
The specific heat capacity of a substance is the quan¬
tity of heat energy required to raise the temperature of
1 kg of the substance by 1°C. The symbol used for spe¬
cific heat capacity is c and the units are J/(kg °C) or
J/(kg K). (Note that these units may also be written as
J kg- 1 “C- 1 or J kg- 1 K 1 .)
Some typical values of specific heat capacity for the
range of temperature 0°C to 100°C include:
Water
Ice
Aluminium
Copper
Iron
Lead
4190 J/(kg °C)
2100 J/(kg °C)
950 J/(kg °C)
390 J/(kg °C)
500 J/(kg °C)
130 J/(kg °C)
Hence to raise the temperature of 1 kg of iron by 1°C
requires 500 J of energy, to raise the temperature of
5 kg of iron by 1°C requires (500 x 5) J of energy, and
to raise the temperature of 5 kg of iron by 40°C re¬
quires (500 x 5 x 40) J of energy, i.e. 100 kJ.
In general, the quantity of heat energy, Q , required
to raise a mass m kg of a substance with a specific heat
capacity of c J/(kg °C), from temperature t { °C to t 2 °C
is given by:
Calculate the quantity of heat
required to raise the temperature of 5 kg of water
from 0°C to 100°C. Assume the specific heat ca¬
pacity of water is 4200 J/(kg °C).
Quantity of heat energy,
Q = mc(t 2 -t l )
= 5 kg x 4200 J/(kg °C) x (100 - 0)°C
= 5 x 4200 x 100
= 2100000 J or 2100 kJ or 2.1 MJ
A block of cast iron having a mass of
10 kg cools from a temperature of 150°C to 50°C.
How much energy is lost by the cast iron? Assume
the specific heat capacity of iron is 500 J/(kg °C).
Quantity of heat energy,
Q = mc(t 2 -t x )
= 10 kg x 500 J/(kg °C) x (50 - 150)°C
= 10 x 500 x (-100)
- -500000 J or -500 kJ or -0.5 MJ
(Note that the minus sign indicates that heat is given
out or lost.)
Some lead having a specific heat
capacity of 130 J/(kg °C) is heated from 27°C to
its melting point at 327°C. If the quantity of heat
required is 780 kJ, determine the mass of the lead.
Quantity of heat, Q = mc(t 2 -1 { ), hence,
780 x 10 3 J = m x 130 J/(kg °C) x (327 - 27)°C
i.e.
780000 = mx 130x300
, 780000 ,
from which, mass, m =-kg = 20 kg
130x 300 6 *
273 kJ of heat energy are required to
raise the temperature of 10 kg of copper from 15°C
to 85°C. Determine the specific heat capacity of
copper.
Quantity of heat, Q = mc(t 2 -1 { ), hence:
273 x 10 3 J = 10 kg x c x (85 - 15)°C
where c is the specific heat capacity,
i.e. 273000 = 10 xcx 70
from which, specific heat capacity, c =
273000
10x70
390 J/(kg °C)
Q = mc(t 2 - t x ) joules
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256 Mechanical Engineering Principles
5.7 MJ of heat energy are supplied to
30 kg of aluminium that is initially at a temperature
of 20°C. If the specific heat capacity of aluminium
is 950 J/(kg°C), determine its final temperature.
Quantity of heat,
Q = mc(t 2 -t x ), hence,
5.7 x 10 6 J = 30 kg x 950 J/(kg °C) x ( t 2 - 20)°C
5.7xl0 6
from which, (t 2 - 20) = 3Q x 95Q = 200
Hence, the final temperature, t 2 = 200 + 20
= 220°C
A copper container of mass 500 g
contains 1 litre of water at 293 K. Calculate the
quantity of heat required to raise the temperature of
the water and container to boiling point, assuming
there are no heat losses. Assume that the specific
heat capacity of copper is 390 J/(kg K), the specific
heat capacity of water is 4.2 kJ/(kg K) and 1 litre of
water has a mass of 1 kg.
Heat is required to raise the temperature of the water,
and also to raise the temperature of the copper container.
For the water: m = 1 kg, t x = 293 K,
t 2 = 373 K (i.e. boiling point) and
c = 4.2 kJ/(kg K)
Quantity of heat required for the water is given by:
/
Q w = mc(t 2 - t { ) = (1 kg)
4.2
kJ
\
V
kgK
(373-293) K
= 4.2 x 80 kJ
i.e. Q w = 336 kJ
For the copper container: m = 500 g = 0.5 kg,
t x = 293 K,
t 2 = 373 K and
c — 390 J/(kg K)
= 0.39 kJ/(kg K)
Quantity of heat required for the copper container is
given by:
Q c = mc(t 2 - t x ) = (0.5 kg)(0.39 kJ/(kg K)(80 K)
i.e. Q c = 15.6 kJ
Total quantity of heat required, Q = Q w + Q c
= 336+ 15.6
= 351.6 kJ
Now try the following Practice Exercise
Practice Exercise 117 Further problems
on specific heat
capacity
1. Determine the quantity of heat energy
(in megajoules) required to raise the tem¬
perature of 10 kg of water from 0°C to 50°C.
Assume the specific heat capacity of water
is 4200 J/(kg °C). [2.1 MJ]
2. Some copper, having a mass of 20 kg,
cools from a temperature of 120°C to 70°C.
If the specific heat capacity of copper is
390 J/(kg °C), how much heat energy is lost
by the copper ? [390 kJ]
3. A block of aluminium having a specific
heat capacity of 950 J/(kg °C) is heated
from 60°C to its melting point at 660°C. If
the quantity of heat required is 2.85 MJ, de¬
termine the mass of the aluminium block.
[5 kg]
4. 20.8 kJ of heat energy is required to raise
the temperature of 2 kg of lead from 16°C to
96°C. Determine the specific heat capacity
of lead. [130 J/(kg°C)]
5. 250 kJ of heat energy is supplied to 10 kg
of iron which is initially at a temperature of
15°C. If the specific heat capacity of iron is
500 J/(kg °C) determine its final tempera¬
ture. [65°C]
21.4 Change of state
A material may exist in any one of three states - solid,
liquid or gas. If heat is supplied at a constant rate to
some ice initially at, say, -30°C, its temperature rises
as shown in Figure 21.1. Initially the temperature in¬
creases from -30°C to 0°C as shown by the line AB. It
then remains constant at 0°C for the time BC required
for the ice to melt into water.
When melting commences the energy gained by
continual heating is offset by the energy required for
the change of state and the temperature remains con¬
stant even though heating is continued. When the ice
is completely melted to water, continual heating raises
the temperature to 100°C, as shown by CD in Figure
21.1. The water then begins to boil and the temperature
Heat energy and transfer 257
again remains constant at 100°C, shown as DE , until all
the water has vaporised.
Continual heating raises the temperature of the
steam as shown by EF in the region where the steam is
termed superheated.
Changes of state from solid to liquid or liquid to gas
occur without change of temperature and such changes
are reversible processes. When heat energy flows to or
from a substance and causes a change of temperature,
such as between A and B , between C and D and be¬
tween E and F in Figure 21.1, it is called sensible heat
(since it can be ‘sensed’ by a thermometer).
Heat energy which flows to or from a substance
while the temperature remains constant, such as be¬
tween B and C and between D and E in Figure 21.1, is
called latent heat (latent means concealed or hidden).
Steam initially at a temperature of
130°C is cooled to a temperature of 20°C below
the freezing point of water, the loss of heat energy
being at a constant rate. Make a sketch, and briefly
explain, the expected temperature/time graph rep¬
resenting this change.
A temperature/time graph representing the change
is shown in Figure 21.2. Initially steam cools until it
reaches the boiling point of water at 100°C. Tempera¬
ture then remains constant, i.e. between^ and B , even
though it is still giving off heat (i.e. latent heat). When
all the steam at 100°C has changed to water at 100°C
it starts to cool again until it reaches the freezing point
of water at 0°C. From C to D the temperature again
remains constant (i.e. latent heat), until all the water
is converted to ice. The temperature of the ice then
decreases as shown.
140
Figure 21.2
Now try the following Practice Exercise
Practice Exercise 118 A further problem
on change of state
1. Some ice, initially at - 40°C, has heat sup¬
plied to it at a constant rate until it becomes
superheated steam at 150°C. Sketch a typi¬
cal temperature/time graph expected and use
it to explain the difference between sensible
and latent heat.
[Similar to Figure 21.1]
21.5 Latent heats of fusion and
vaporisation
The specific latent heat of fusion is the heat required
to change 1 kg of a substance from the solid state to
the liquid state (or vice versa) at constant temperature.
The specific latent heat of vaporisation is the heat re¬
quired to change 1 kg of a substance from a liquid to a
gaseous state (or vice versa) at constant temperature.
The units of the specific latent heats of fusion and
vaporisation are J/kg, or more often kJ/kg, and some
typical values are shown in Table 21.1 on page 258.
The quantity of heat Q supplied or given out during a
change of state is given by:
Q = mL
where m is the mass in kilograms and L is the specific
latent heat.
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258 Mechanical Engineering Principles
Table 21.1
Latent heat
of fusion
(w/kg)
Melting
point (°C)
Mercury
11.8
-39
Lead
22
327
Silver
100
957
Ice
335
0
Aluminium
387
660
Latent heat of
Boiling
vaporisation
(w/kg)
point (°C)
Oxygen
214
- 183
Mercury
286
357
Ethyl alcohol
857
79
Water
2257
100
Thus, for example, the heat required to convert 10 kg
of ice at 0°C to water at 0°C is given by
10 kg x 335 kJ/kg - 3350 kJ or 3.35 MJ
Besides changing temperature, the effects of supplying
heat to a material can involve changes in dimensions,
as well as in colour, state and electrical resistance.
Most substances expand when heated and contract
when cooled, and there are many practical applications
and design implications of thermal movement (see
Chapter 22).
How much heat is needed to melt
completely 12 kg of ice at 0°C ? Assume the latent
heat of fusion of ice is 335 kJ/kg.
Quantity of heat required, Q = mL = 12 kg x 335 kJ/kg
= 4020 kJ or 4.02 MJ
Calculate the heat required to con¬
vert 5 kg of water at 100°C to superheated steam
at 100°C. Assume the latent heat of vaporisation of
water is 2260 kJ/kg.
Quantity of heat required, Q = mL = 5 kg x 2260 kJ/kg
= 11300 kJ or 11.3 MJ
Determine the heat energy needed
to convert 5 kg of ice initially at - 20°C completely
to water at 0°C. Assume the specific heat capacity
of ice is 2100 J/(kg °C) and the specific latent heat
of fusion of ice is 335 kJ/kg.
Quantity of heat energy needed, Q = sensible heat +
latent heat.
The quantity of heat needed to raise the temperature of
ice from -20°C to 0°C, i.e. sensible heat,
Qx =mc{t 2 -t x )
= 5 kg x 2100 J/(kg°C) x (0 - -20)°C
= (5 x2100x20) J = 210 kJ
The quantity of heat needed to melt 5 kg of ice at 0°C,
i.e. the latent heat, Q 2 = mL
= 5 kg x 335 kJ/kg
=1675 kJ
Total heat energy needed, Q = Q x + e 2
= 210+ 1675 = 1885 kJ
Calculate the heat energy required
to convert completely 10 kg of water at 50°C into
steam at 100°C, given that the specific heat capac¬
ity of water is 4200 J/(kg °C) and the specific latent
heat of vaporisation of water is 2260 kJ/kg.
Quantity of heat required = sensible heat + latent heat.
Sensible heat,
Q x =mc(t 2 -t x )
= 10 kg x 4200 J/(kg °C) x (100 - 50)°C
= 2100 kJ
Latent heat, Q 2 = mL = 10 kg x 2260 kJ/kg
= 22600 kJ
Total heat energy required, Q = Q x + Q 2
= (2100 + 22600) kJ
= 24700 kJ or 24.70 MJ
Determine the amount of heat
energy needed to change 400 g of ice, initially at
-20°C, into steam at 120°C. Assume the following:
latent heat of fusion of ice = 335 kJ/kg, latent heat
of vaporisation of water = 2260 kJ/kg, specific
heat capacity of ice = 2.14 kJ/(kg °C), specific heat
capacity of water = 4.2 kJ/(kg °C) and specific heat
capacity of steam = 2.01 kJ/(kg °C).
Heat energy and transfer 259
The energy needed is determined in five stages:
(i) Heat energy needed to change the temperature of
ice from - 20°C to 0°C is given by:
Qi = mc(t 2 - t { )
= 0.4 kg x 2.14 kJ/(kg °C) x (0 —20)°C
= 17.12 kJ
(ii) Latent heat needed to change ice at 0°C into
water at 0°C is given by:
Q 2 = mL f = 0.4 kg x 335 kJ/kg = 134 kJ
(iii) Heat energy needed to change the temperature
of water from 0°C (i.e. melting point) to 100°C
(i.e. boiling point) is given by:
Q 3 = mc(t 2 - t { )
= 0.4 kg x 4.2 kJ/(kg °C) x 100°C
= 168 kJ
(iv) Latent heat needed to change water at 100°C into
steam at 100°C is given by:
Q 4 = mL v = 0.4 kg x 2260 kJ/kg - 904 kJ
(v) Heat energy needed to change steam at 100°C
into steam at 120°C is given by:
Q 5 = mc(t [ - t 2 )
= 0.4 kg x 2.01 kJ/(kg °C) x 20°C
= 16.08 kJ
Total heat energy needed,
Q = Q\ + 02 + 03 + 04 + ft
= 17.12 + 134 + 168 + 904 + 16.08
= 1239.2 kJ
Now try the following Practice Exercise
Practice Exercise 119 Further problems
on the latent heats
of fusion and
vaporisation
1. How much heat is needed to melt completely
25 kg of ice at 0°C. Assume the specific
latent heat of fusion of ice is 335 kJ/kg.
[8.375 MJ]
2. Determine the heat energy required to change
8 kg of water at 100°C to superheated steam
at 100°C. Assume the specific latent heat of
vaporisation of water is 2260 kJ/kg.
[18.08 MJ]
3. Calculate the heat energy required to convert
10 kg of ice initially at - 30°C completely
into water at 0°C. Assume the specific heat
capacity of ice is 2.1 kJ/(kg °C) and the spe¬
cific latent heat of fusion of ice is 335 kJ/kg.
[3.98 MJ]
4. Determine the heat energy needed to convert
completely 5 kg of water at 60°C to steam
at 100°C, given that the specific heat ca¬
pacity of water is 4.2 kJ/(kg °C) and the
specific latent heat of vaporisation of water is
2260 kJ/kg. [12.14 MJ]
21.6 A simple refrigerator
The boiling point of most liquids may be lowered if
the pressure is lowered. In a simple refrigerator a
working fluid, such as ammonia or freon, has the
pressure acting on it reduced. The resulting lower¬
ing of the boiling point causes the liquid to vaporise.
In vaporising, the liquid takes in the necessary latent
heat from its surroundings, i.e. the freezer, which
thus becomes cooled. The vapour is immediately re¬
moved by a pump to a condenser that is outside of the
cabinet, where it is compressed and changed back into
a liquid, giving out latent heat. The cycle is repeated
when the liquid is pumped back to the freezer to be
vaporised.
21.7 Conduction, convection
and radiation
Heat may be transferred from a hot body to a
cooler body by one or more of three methods, these
being: (a) by conduction (b) by convection or (c) by
radiation.
Conduction
Conduction is the transfer of heat energy from one
part of a body to another (or from one body to another)
without the particles of the body moving.
Conduction is associated with solids. For example,
if one end of a metal bar is heated, the other end will
become hot by conduction. Metals and metallic alloys
are good conductors of heat, whereas air, wood, plastic,
cork, glass and gases are examples of poor conductors
(i.e. they are heat insulators).
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260 Mechanical Engineering Principles
Practical applications of conduction include:
(i) A domestic saucepan or dish conducts heat from
the source to the contents. Also, since wood and
plastic are poor conductors of heat they are used
for saucepan handles.
(ii) The metal of a radiator of a central heating sys¬
tem conducts heat from the hot water inside to
the air outside.
Convection
Convection is the transfer of heat energy through a
substance by the actual movement of the substance it¬
self. Convection occurs in liquids and gases, but not
in solids. When heated, a liquid or gas becomes less
dense. It then rises and is replaced by a colder liquid
or gas and the process repeats. For example, electric
kettles and central heating radiators always heat up at
the top first.
Examples of convection are:
(i) Natural circulation hot water heating systems
depend on the hot water rising by convection to
the top of the house and then falling back to the
bottom of the house as it cools, releasing the heat
energy to warm the house as it does so.
(ii) Convection currents cause air to move and there¬
fore affect climate.
(iii) When a radiator heats the air around it, the hot
air rises by convection and cold air moves in to
take its place.
(iv) A cooling system in a car radiator relies on con¬
vection.
(v) Large electrical transformers dissipate waste
heat to an oil tank. The heated oil rises by con¬
vection to the top, then sinks through cooling
fins, losing heat as it does so.
(vi) In a refrigerator, the cooling unit is situated near
the top. The air surrounding the cold pipes be¬
come heavier as it contracts and sinks towards
the bottom. Warmer, less dense air is pushed
upwards and in turn is cooled. A cold convection
current is thus created.
Radiation
Radiation is the transfer of heat energy from a hot
body to a cooler one by electromagnetic waves. Heat
radiation is similar in character to light waves - it
travels at the same speed and can pass through a
vacuum - except that the frequency of the waves is
different. Waves are emitted by a hot body, are trans¬
mitted through space (even a vacuum) and are not
detected until they fall on to another body. Radiation
is reflected from shining, polished surfaces but ab¬
sorbed by dull, black surfaces.
Practical applications of radiation include:
(i) heat from the sun reaching earth
(ii) heat felt by a flame
(iii) cooker grills
(iv) industrial furnaces
(v) infra-red space heaters.
21.8 Vacuum flask
A cross-section of a typical vacuum flask is shown in
Figure 21.3 and is seen to be a double-walled bottle
with a vacuum space between them, the whole sup¬
ported in a protective outer case.
Cork stopper
Figure 21.3
Very little heat can be transferred by conduction be¬
cause of the vacuum space and the cork stopper (cork is
a bad conductor of heat). Also, because of the vacuum
space, no convection is possible. Radiation is mini¬
mised by silvering the two glass surfaces (radiation is
reflected off shining surfaces).
Thus a vacuum flask is an example of prevention of
all three types of heat transfer and is therefore able to
keep hot liquids hot and cold liquids cold.
21.9 Use of insulation in
conserving fuel
Fuel used for heating a building is becoming increas¬
ingly expensive. By the careful use of insulation, heat
can be retained in a building for longer periods and the
cost of heating thus minimised.
Heat energy and transfer 261
(i) Since convection causes hot air to rise it is im¬
portant to insulate the roof space, which is prob¬
ably the greatest source of heat loss in the home.
This can be achieved by laying fibre-glass be¬
tween the wooden joists in the roof space.
(ii) Glass is a poor conductor of heat. However,
large losses can occur through thin panes of glass
and such losses can be reduced by using double-
glazing. Two sheets of glass, separated by air,
are used. Air is a very good insulator but the air
space must not be too large otherwise convec¬
tion currents can occur which would carry heat
across the space.
(iii) Hot water tanks should be lagged to prevent con¬
duction and convection of heat to the surround¬
ing air.
(iv) Brick, concrete, plaster and wood are all poor
conductors of heat. A house is made from two
walls with an air gap between them. Air is a
poor conductor and trapped air minimises losses
through the wall. Heat losses through the walls
can be prevented almost completely by using
cavity wall insulation, i.e. plastic-foam.
Besides changing temperature, the effects of supplying
heat to a material can involve changes in dimensions, as
well as in colour, state and electrical resistance.
Most substances expand when heated and contract
when cooled, and there are many practical applications
and design implications of thermal movement as ex¬
plained in Chapter 22 following.
Now try the following Practice Exercises
Practice Exercise 120 Short-answer
questions on
heat energy
1. Differentiate between temperature and
heat.
2. Name two scales on which temperature is
measured.
3. Name any four temperature measuring
devices.
4. Define specific heat capacity and name its
unit.
5. Differentiate between sensible and latent
heat.
6. The quantity of heat, 0, required to raise a
mass m kg from temperature t° C to t 2 °C,
the specific heat capacity being c, is given
by Q = .
7. What is meant by the specific latent heat of
fusion?
8. Define the specific latent heat of vaporisa¬
tion.
9. Explain briefly the principle of operation of
a simple refrigerator.
10. State three methods of heat transfer.
11. Define conduction and state two practical
examples of heat transfer by this method.
12. Define convection and give three examples
of heat transfer by this method.
13. What is meant by radiation? Give three
uses.
14. How can insulation conserve fuel in a typi¬
cal house?
Practice Exercise 121 Multiple-choice
questions on heat
energy
(Answers on page 336)
1. Heat energy is measured in:
(a) kelvin (b) watts
(c) kilograms (d) joules
2. A change of temperature of 20°C is equiva¬
lent to a change in thermodynamic temper¬
ature of:
(a) 293 K (b) 20 K
(c) 80 K (d) 120 K
3. A temperature of 20°C is equivalent to:
(a) 293 K (b) 20 K
(c) 80 K (d) 120 K
4. The unit of specific heat capacity is:
(a) joules per kilogram
(b) joules
(c) joules per kilogram kelvin
(d) cubic metres
5. The quantity of heat required to raise
the temperature of 500 g of iron by 2°C,
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262 Mechanical Engineering Principles
given that the specific heat capacity is
500 J/(kg °C), is:
(a) 500 kJ (b) 0.5 kJ
(c) 2 J (d) 250 kJ
6. The heat energy required to change 1 kg of
a substance from a liquid to a gaseous state
at the same temperature is called:
(a) specific heat capacity
(b) specific latent heat of vaporisation
(c) sensible heat
(d) specific latent heat of fusion
7. The temperature of pure melting ice is:
(a) 373 K (b) 273 K
(c) 100 K (d) OK
8. 1.95 kJ of heat is required to raise the tem¬
perature of 500 g of lead from 15°C to its
final temperature. Taking the specific heat
capacity of lead to be 130 J/(kg °C), the
final temperature is:
(a) 45°C (b) 37.5°C
(c) 30°C (d) 22.5°C
9. Which of the following temperature is ab¬
solute zero ?
(a) 0°C (b) - 173°C
(c) - 273°C (d) - 373°C
10. When two wires of different metals are
twisted together and heat applied to the
junction, an e.m.f. is produced. This effect
is used in a thermocouple to measure:
(a) e.m.f. (b) temperature
(c) expansion (d) heat
11. Which of the following statements is false?
(a) - 30°C is equivalent to 243 K.
(b) Convection only occurs in liquids and
gases.
(c) Conduction and convection cannot
occur in a vacuum.
(d) Radiation is absorbed by a silver
surface.
12. The transfer of heat through a substance by
the actual movement of the particles of the
substance is called:
(a) conduction
(b) radiation
(c) convection
(d) specific heat capacity
13. Which of the following statements is
true?
(a) Heat is the degree of hotness or cold¬
ness of a body.
(b) Heat energy that flows to or from a sub¬
stance while the temperature remains
constant is called sensible heat.
(c) The unit of specific latent heat of fusion
is J/(kg K).
(d) A cooker-grill is a practical application
of radiation.
For fully worked solutions to each of the problems in Practice Exercises 116 to 121 in this chapter,
go to the website:
www.routledge.com/cw/bird
Chapter 22
Why it is important to understand: Thermal expansion
Thermal expansion and contraction are very important features in engineering science. For example, if the
metal railway lines of a railway track are heated or cooled due to weather conditions or time of day, their
lengths can increase or decrease accordingly. If the metal lines are heated due to the weather effects, then
the railway lines will attempt to expand, and depending on their construction, they can buckle, rendering
the track useless for transporting trains. In countries with large temperature variations, this effect can be
much worse, and the engineer may have to choose a superior metal to withstand these changes. The effect
of metals expanding and contracting due to the rise and fall of temperatures, accordingly, can also be put
to good use. A classic example of this is the simple humble domestic thermostat, which when the water gets
too hot, will cause the metal thermostat to expand and switch off the electric heater; conversely, when the
water becomes too cool, the metal thermostat shrinks, causing the electric heater to switch on again. All
sorts of materials, besides metals, are affected by thermal expansion and contraction. The chapter also
defines the coefficients of Unear, superficial and cubic expansion. Thermal expansion is of importance in
the design and construction of structures, switches and other bodies.
At the end of this chapter you should be able to:
• appreciate that expansion and contraction occurs with change of temperature
• describe practical applications where expansion and contraction must be allowed for
• understand the expansion and contraction of water
• define the coefficient of linear expansion a
• recognise typical values for the coefficient of linear expansion
• calculate the new length L 2 , after expansion or contraction, using L 2 = L { \ 1 + a(t 2 - t { )]
• define the coefficient of superficial expansion /?
• calculate the new surface area^ 2 , after expansion or contraction, using A 2 = A l [l + p(t 2 - fj)]
• appreciate that /? ~ 2 a
• define the coefficient of cubic expansion y
• recognise typical values for the coefficient of cubic expansion
• appreciate that y ~ 3a
• calculate the new volume V 2 , after expansion or contraction, using V 2 = V ] [1 + y(t 2 - ^)]
22.1 Introduction
When heat is applied to most materials, expansion
occurs in all directions (see Section 3.12, page 58).
Conversely, if heat energy is removed from a material
(i.e. the material is cooled) contraction occurs in all
directions. The effects of expansion and contraction each
depend on the change of temperature of the material.
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Four
264 Mechanical Engineering Principles
22.2 Practical applications of thermal
expansion
22.3 Expansion and contraction
of water
Some practical applications where expansion and con¬
traction of solid materials must be allowed for include:
(i) Overhead electrical transmission lines are hung
so that they are slack in summer, otherwise their
contraction in winter may snap the conductors or
bring down pylons.
(ii) Gaps need to be left in lengths of railway lines
to prevent buckling in hot weather (except where
these are continuously welded).
(iii) Ends of large bridges are often supported on roll¬
ers to allow them to expand and contract freely.
(iv) Fitting a metal collar to a shaft or a steel tyre to a
wheel is often achieved by first heating the collar
or tyre so that they expand, fitting them in posi¬
tion, and then cooling them so that the contrac¬
tion holds them firmly in place; this is known as
a ‘shrink-fit’. By a similar method hot rivets are
used for joining metal sheets.
(v) The amount of expansion varies with different
materials. Figure 22.1(a) shows a bimetallic strip
at room temperature (i.e. two different strips
of metal riveted together - see Section 3.12,
page 58). When heated, brass expands more
than steel, and since the two metals are riveted
together the bimetallic strip is forced into an arc
as shown in Figure 22.1(b). Such a movement
can be arranged to make or break an electric
circuit and bimetallic strips are used, in par¬
ticular, in thermostats (which are temperature-
operated switches) used to control central heating
systems, cookers, refrigerators, toasters, irons,
hot-water and alarm systems.
Steel
'(a) (b)
Figure 22.1
(vi) Motor engines use the rapid expansion of heated
gases to force a piston to move.
(vii) Designers must predict, and allow for, the ex¬
pansion of steel pipes in a steam-raising plant
so as to avoid damage and consequent danger to
health.
Water is a liquid that at low temperature displays an un¬
usual effect. If cooled, contraction occurs until, at about
4°C, the volume is at a minimum. As the temperature
is further decreased from 4°C to 0°C expansion occurs,
i.e. the volume increases. (For cold, deep fresh water,
the temperature at the bottom is more likely to be about
4°C, somewhat warmer than less deep water.) When ice
is formed, considerable expansion occurs and it is this
expansion that often causes frozen water pipes to burst.
A practical application of the expansion of a liquid is
with thermometers, where the expansion of a liquid, such
as mercury or alcohol, is used to measure temperature.
22.4 Coefficient of linear expansion
The amount by which unit length of a material expands
when the temperature is raised one degree is called the
coefficient of linear expansion of the material and is rep¬
resented by a (Greek alpha) - see Section 3.12, page 58.
The units of the coefficient of linear expansion are
m/(mK), although it is usually quoted as just K or K _1 . For
example, copper has a coefficient of linear expansion value
of 17 x KF 6 K -1 , which means that aim long bar of cop¬
per expands by 0.000017 m if its temperature is increased
by 1 K (or 1°C). If a 6 m long bar of copper is subjected
to a temperature rise of 25 K then the bar will expand by
(6 x 0.000017 x 25) m, i.e. 0.00255 m or 2.55 mm. (Since
the Kelvin scale uses the same temperature interval as the
Celsius scale, a change of temperature of, say, 50°C, is the
same as a change of temperature of 50 K.)
If a material, initially of length L { and at a tempera¬
ture of t j and having a coefficient of linear expansion
a , has its temperature increased to t 2 , then the new
length L 2 of the material is given by:
New length = original length + expansion
i.e. L 2 = L x + L x a(t 2 -1±)
i.e. L 2 = L x [1 + a(t 2 - t t )] (22.1)
Some typical values for the coefficient of linear expan¬
sion include:
Aluminium
23 x 1(T 6 K 1
Brass
18 x 10 -6 Kr 1
Concrete
12 x 10- 6 Kr 1
Copper
17 x 10 -6 Kr 1
Gold
14 x 10 6 K 1
Invar
(nickel-
steel
alloy)
0.9 x 10- 6 K 1
Thermal expansion 265
Iron 11 -12 x Nylon lOOxlO^Kr 1
10- 6 K 1
Steel 15 - 16 x Tungsten 4.5 x lO -6 Kr 1
10 6 K 1
Zinc 31 x10- 6 K-*
The length of an iron steam pipe is
20.0 m at a temperature of 18°C. Determine the
length of the pipe under working conditions when
the temperature is 300°C. Assume the coefficient of
linear expansion of iron is 12 x 10 6 K -1 .
Length L { = 20.0 m, temperature t { = 18°C,
t 2 = 300°C and a = 12 x 10" 6 K" 1
Length of pipe at 300°C is given by:
I*2 ~ [1 Cf.(t 2 ~ ^)]
= 20.0[1 + (12 x 10- 6 )(300- 18)]
= 20.0[1 + 0.003384]
- 20.0[1.003384] - 20.06768 m
i.e. an increase in length of 0.06768 m or 67.68 mm.
In practice, allowances are made for such expansions.
U-shaped expansion joints are connected into pipelines
carrying hot fluids to allow some ‘give’ to take up the
expansion.
An electrical overhead transmission
line has a length of 80.0 m between its supports
at 15°C. Its length increases by 92 mm at 65°C.
Determine the coefficient of linear expansion of the
material of the line.
Length L x = 80.0 m, L 1 = 80.0 m + 92 mm
= 80.092 m,
temperature t { = 15°C and temperature t 2 = 65°C
Length L 2 = L { [1 + a(t 2 -t { )]
i.e. 80.092 = 80.0[1 + a(65- 15)]
80.092 = 80.0 + (80.0)(a)(50)
i.e. 80.092 - 80.0 = (80.0)(a)(50)
Hence, the coefficient of linear expansion,
0.092
a= (800X50) =0 - 000023
i.e. a = 23 x 10 -6 K -1 (which is aluminium - see above)
A measuring tape made of copper
measures 5.0 m at a temperature of 288 K.
Calculate the percentage error in measurement
when the temperature has increased to 313 K.
Take the coefficient of linear expansion of copper
as 17 x 10 -6 Kr 1 .
Length L l = 5.0 m, temperature t { =288 K,
t 2 = 313 K and cc = 17 x 10“ 6 K -1
Length at 313 K is given by:
Length Z, 9 = L x [1 + a(t 2 - 1 { )]
= 5.0[1 + (17 x 10 _6 )(313 - 288)]
= 5.0[1 + (17 x 10 _6 )(25)]
= 5.0[1 +0.000425]
= 5.0[1.000425] = 5.002125 m
i.e. the length of the tape has increased by 0.002125 m
Percentage error in measurement at 313 K
increase in length
original length
x 100%
0.002125
———x 100 = 0.0425%
The copper tubes in a boiler are 4.20 m
long at a temperature of 20°C. Determine the
length of the tubes (a) when surrounded only by
feed water at 10°C (b) when the boiler is operating
and the mean temperature of the tubes is 320°C.
Assume the coefficient of linear expansion of
copper to be 17 x 10~ 6 K 1 .
(a) Initial length, L x = 4.20 m, initial temperature,
t x = 20°C, final temperature, t 2 = 10°C and
a = 17 x 10 -6 Kr 1
Final length at 10°C is given by:
Z>2 — Lj [1 + a(t 2 — ^)]
= 4.20[1 + (17 x 10 -6 )(10 - 20)]
= 4.20[1 -0.00017]= 4.1993 m
i.e. the tube contracts by 0.7 mm when the
temperature decreases from 20°C to 10°C
(b) Length, L { = 4.20 m, t { = 20°C, t 2 = 320°C and
a= 17 x 10 -6 K -1
Final length at 320°C is given by:
L 2 — L x [1 + a(t 2 — ^)]
= 4.20[1 + (17 x 10" 6 )(320 - 20)]
= 4.20[1 +0.0051] =4.2214 m
Part Four
Part Four
266 Mechanical Engineering Principles
i.e. the tube extends by 21.4 mm when the
temperature rises from 20°C to 320°C
Now try the following Practice Exercise
Practice Exercise 122 Further problems
on the coefficient of
linear expansion
1. A length of lead piping is 50.0 m long at a
temperature of 16°C. When hot water flows
through it the temperature of the pipe rises to
80°C. Determine the length of the hot pipe if
the coefficient of linear expansion of lead is
29 x 10~ 6 K _1 . [50.0928 m]
2. A rod of metal is measured at 285 K and is
3.521 m long. At 373 K the rod is 3.523 m
long. Determine the value of the coefficient
of linear expansion for the metal.
[6.45 x 10 -6 K -1 ]
3. A copper overhead transmission line has
a length of 40.0 m between its supports at
20°C. Determine the increase in length at
50°C if the coefficient of linear expansion of
copper is 17 x 10 -6 K 1 . [20.4 mm]
4. A brass measuring tape measures 2.10 m at a
temperature of 15°C. Determine
(a) the increase in length when the tempera¬
ture has increased to 40°C
(b) the percentage error in measurement at
40°C. Assume the coefficient of linear
expansion of brass to be 18 x 10 -6 K _1 .
[(a) 0.945 mm (b) 0.045%]
5. A pendulum of a ‘grandfather’ clock is
2.0 m long and made of steel. Determine
the change in length of the pendulum if
the temperature rises by 15 K. Assume the
coefficient of linear expansion of steel to be
15 x 10 -6 K _1 . [0.45 mm]
6. A temperature control system is operated by
the expansion of a zinc rod which is 200 mm
long at 15°C. If the system is set so that the
source of heat supply is cut off when the rod
has expanded by 0.20 mm, determine the
temperature to which the system is limited.
Assume the coefficient of linear expansion
of zinc to be 31 x 10" 6 K _1 . [47.26°C]
7. A length of steel railway line is 30.0 m
long when the temperature is 288 K. Deter¬
mine the increase in length of the line when
the temperature is raised to 303 K. Assume
the coefficient of linear expansion of steel to
be 15 x 10 -6 K _1 . [6.75 mm]
8. A brass shaft is 15.02 mm in diameter and
has to be inserted in a hole of diameter
15.0 mm. Determine by how much the shaft
must be cooled to make this possible, with¬
out using force. Take the coefficient of linear
expansion of brass as 18 x 10~ 6 K _1 . [74 K]
22.5 Coefficient of superficial
expansion
The amount by which unit area of a material increases
when the temperature is raised by one degree is called
the coefficient of superficial (i.e. area) expansion and
is represented by ft (Greek beta).
If a material having an initial surface area A l at
temperature t ] and having a coefficient of superficial
expansion /?, has its temperature increased to t 2 , then
the new surface area A 2 of the material is given by:
New surface area = original surface area + increase
in area
i.e. A 2 = A { + A x P(t 2 -t { )
i.e. A 2 =A, [1 +li(t 2 -t x )\ (22.2)
It is shown in Problem 5 below that the coefficient of
superficial expansion is twice the coefficient of linear
expansion, i.e. ft = 2a , to a very close approximation.
Show that for a rectangular area of
material having dimensions L by b, the coefficient
of superficial expansion ft ~ 2a, where a is the
coefficient of linear expansion.
Initial area, A l = Lb. For a temperature rise of 1 K,
side L will expand to (L + La) and side b will expand
to ( b + ba). Hence the new area of the rectangle, A 2 , is
given by:
A 2 = (L + La)(b + ba)
= L{ 1 + a)b( 1 + a) = Lb( 1 + a) 2
= Lb{ 1 + 2a + a 2 ) ~ Lb( 1 + 2a)
since a 2 is very small (see typical values in Section 22.4)
Thermal expansion 267
Hence, A 2 ~A l (1 + 2a)
For a temperature rise of ( t 2 — t x ) K
A 2 ~A x [1 + la{t 2 - fj)]
Thus, from equation (22.2), p~2a
22.6 Coefficient of cubic expansion
The amount by which unit volume of a material
increases for a one degree rise of temperature is called
the coefficient of cubic (or volumetric) expansion
and is represented by y (Greek gamma).
If a material having an initial volume V x at tem¬
perature t x and having a coefficient of cubic expansion
y, has its temperature raised to t 2 , then the new volume
of the material is given by:
New volume = initial volume + increase in volume
i.e. F 2 = F, + F, y(t 2 -t 1 )
i.e. \\+y(t 2 -t,)\ (22.3)
It is shown in Problem 6 below that the coefficient of
cubic expansion is three times the coefficient of linear
expansion, i.e. y = 3a, to a very close approximation.
A liquid has no definite shape and only its cubic or
volumetric expansion need be considered. Thus with
expansions in liquids, equation (22.3) is used.
Show that for a rectangular block
of material having dimensions T, b and /z, the
coefficient of cubic expansion y ~ 3a, where a
is the coefficient of linear expansion.
Initial volume, V 1 = Lbh. For a temperature rise of 1 K,
side L expands to (L + La ), side b expands to (b + ba)
and side h expands to (h + ha)
Hence the new volume of the block V 2 is given by:
V 2 = (L + La)(b + ba){h + ha)
= L(l + a)b( 1 + a)h{ 1 + a)
= Lbh{\ + a) 3 = Lbh{ 1 + 3a + 3 a 2 + a 3 )
~Lbh(l +3a)
since terms in a 2 and a 3 are very small
Hence, F 2 ~E 1 (l+3a)
For a temperature rise of (t 2 — t x ) K,
F 2 ^[l+3 a(t 2 -t x )\
Thus, from equation (22.3), y~3a
Some typical values for the coefficient of cubic expan¬
sion measured at 20°C (i.e. 293 K) include:
Ethyl 1.1 x 10" 3 K" 1 Mercury 1.82 x 1(H K" 1
alcohol
Paraffin 9 x 10" 2 K" 1 Water 2.1 x 10" 4 K -1
oil
The coefficient of cubic expansion y is only constant
over a limited range of temperature.
A brass sphere has a diameter of
50 mm at a temperature of 289 K. If the temperature
of the sphere is raised to 789 K, determine the
increase in (a) the diameter (b) the surface area
(c) the volume of the sphere. Assume the coefficient
of linear expansion for brass is 18 x 10 -6 K -1 .
(a) Initial diameter, L x = 50 mm, initial temperature,
t x = 289 K, final temperature, t 2 = 789 K and
a = 18 x 10 -6 K -1 .
New diameter at 789 K is given by:
L 2 =L x [1 +a(t 2 -t x )\ from equation (22.1)
i.e. L 2 = 50[1 + (18 x 10" 6 )(789 - 289)]
= 50[1 +0.009] = 50.45 mm
Hence the increase in the diameter is 0.45 mm
Initial surface area of sphere,
A 1 = 4zrr 2 = 4k
r \2
50
v
= 25007T mm 2
7
New surface area at 789 K is given by:
A 1 =A X [ 1 + P(t 2 - fj)] from equation (22.2)
i.e. A 2 =A x [1 + la(t 2 - fj)]
since ft = 2a, to a very close approximation
Thus A 2 = 2500tt[1 + 2(18 x 10- 6 )(500)]
= 25007t[1 + 0.018]
= 25007T + 25007t(0.018)
Hence increase in surface area = 25 007r(0.018)
= 141.4 mm 2
4
Initial volume of sphere, V x = — Kr
r \3
50
K
v 2 .
mnr
New volume at 789 K is given by:
V 2 = V x [\ + y(t 2 - tj)] from equation (22.3)
i.e. V 2 = V { [1 + 3 a(t 2 - ^)]
Part Four
Part Four
268 Mechanical Engineering Principles
since y = 3 a, to a very close approximation
Thus V 2 = -tt( 25) 3 [1 + 3(18 x 10 6 )(500)]
3
= — 7r(25) 3 [1 +0.027]
3
= — 7t(25) 3 + — 7t(25) 3 (0.027)
3 3
4 i
Hence, the increase in volume = — 7r(25) J (0.027)
3
= 1767 mm 3
Mercury contained in a thermometer
has a volume of 476 mm 3 at 15°C. Determine the
temperature at which the volume of mercury is
478 mm 3 , assuming the coefficient of cubic
expansion for mercury to be 1.8 x 10 -4 K -1 .
Initial volume, V l = 476 mm 3 , final volume,
V 2 = 478 mm 3 , initial temperature, t { = 15°C and
7 = 1.8 x lO- 4 * * K- 1
Final volume, V 2 = V { [1 + y(t 2 - ^)]
from equation (22.3)
i.e. V 2 =V \ + V i y( ( 2 ~ f i)
V 2 -V l
from which, (t 2 - fj) = —y ^
_ 478 - 476
~ (476)(1.8xl0-4)
= 23.34°C
Hence, t 2 = 23.34 + t { = 23.34 + 15 = 38.34°C
Hence, the temperature at which the volume of
mercury is 478 mm 3 is 38.34°C
A rectangular glass block has a length
of 100 mm, width 50 mm and depth 20 mm at
293 K. When heated to 353 K its length increases
by 0.054 mm. What is the coefficient of linear
expansion of the glass? Find also (a) the increase
in surface area (b) the change in volume resulting
from the change of length.
Final length, L 2 = L x [1 + a{t 2 - ^)]
from equation (22.1),
hence increase in length is given by:
L 2 — L x = L { a(t 2 - fj)
Hence 0.054 = (100)(a)(353 - 293)
from which, the coefficient of linear expansion is
given by:
a
0.054
(100X60)
= 9 x 10 6 K 1
(a) Initial surface area of glass,
A l = (2 x 100 x 50) + (2 x 50 x 20)
+ (2 x 100x20)
= 10000 + 2000 + 4000 = 16000 mm 2
Final surface area of glass (from equation (22.2)),
A 2 — A j [1 + fi(t 2 — /j)] — A j [1 + 2a(t 2 — ^)]
since /? = 2a to a very close approximation
Hence, increase in surface area
= A X (2a)(? 2 -? 1 ) = (16000)(2x9x 10 6 )(60)
= 17.28 mm 2
(b) Initial volume of glass, V 1 = 100 x 50 x 20
= 100000 mm 3
Final volume of glass (from equation (22.3)),
V 2 = V 1 [l+y(t 2 -t l )]
= F, [1 +3a(t 2 -t l )]
since y = 3a to a very close approximation
Hence, increase in volume of glass
= v i (3a)(t 2 - ?,)
= (100000)(3 x 9 x 10- 6 )(60) = 162 mm 3
Now try the following Practice Exercises
Practice Exercise 123 Further problems on
the coefficients of
superficial and cubic
expansion
1. A silver plate has an area of 800 mm 2 at
15°C. Determine the increase in the area
of the plate when the temperature is raised
to 100°C. Assume the coefficient of linear
expansion of silver to be 19 x 10~ 6 K 1 .
[2.584 mm 2 ]
2. At 283 K a thermometer contains 440 mm 3
of alcohol. Determine the temperature at
which the volume is 480 mm 3 assuming
that the coefficient of cubic expansion of the
alcohol is 12 x 10" 4 K -1 . [358.8 K]
3. A zinc sphere has a radius of 30.0 mm at a
temperature of 20°C. If the temperature of
the sphere is raised to 420°C, determine the
Thermal expansion 269
increase in: (a) the radius, (b) the surface
area, (c) the volume of the sphere. Assume
the coefficient of linear expansion for zinc to
be 31 x 10 6 K
[(a) 0.372 mm (b) 280.5 mm 2 (c) 4207 mm 3 ]
4. A block of cast iron has dimensions of 50 mm
by 30 mm by 10 mm at 15°C. Determine the
increase in volume when the temperature of
the block is raised to 75°C. Assume the coef¬
ficient of linear expansion of cast iron to be
11 x 10~ 6 K _1 . [29.7 mm 3 ]
5. Two litres of water, initially at 20°C, is
heated to 40°C. Determine the volume of
water at 40°C if the coefficient of volumet¬
ric expansion of water within this range is
30 x 10~ 5 K 3 . [2.012 litres]
6. Determine the increase in volume, in litres,
of 3 m 3 of water when heated from 293 K to
boiling point if the coefficient of cubic ex¬
pansion is 2.1 x lCT 4 K -1 (1 litre ~ 10 -3 m 3 ).
[50.4 litres]
7. Determine the reduction in volume when the
temperature of 0.5 litre of ethyl alcohol is
reduced from 40°C to -15°C. Take the coef¬
ficient of cubic expansion for ethyl alcohol
as 1.1 x 10- 3 K" 1 . [0.03025 litres]
Practice Exercise 124 Short-answer
questions on
thermal expansion
1. When heat is applied to most solids and
liquids.occurs.
2. When solids and liquids are cooled they
usually.
3. State three practical applications where the
expansion of metals must be allowed for.
4. State a practical disadvantage where the
expansion of metals occurs.
5. State one practical advantage of the expan¬
sion of liquids.
6. What is meant by the ‘coefficient of expan¬
sion’?
7. State the symbol and the unit used for the
coefficient of linear expansion.
8. Define the ‘coefficient of superficial expan¬
sion’ and state its symbol.
9. Describe how water displays an unexpect¬
ed effect between 0°C and 4°C.
10. Define the ‘coefficient of cubic expansion’
and state its symbol.
Practice Exercise 125 Multiple-choice
questions on
thermal expansion
(Answers on page 336)
1. When the temperature of a rod of copper is
increased, its length:
(a) stays the same (b) increases
(c) decreases
2. The amount by which unit length of a ma¬
terial increases when the temperature is
raised one degree is called the coefficient of:
(a) cubic expansion
(b) superficial expansion
(c) linear expansion
3. The symbol used for volumetric expansion
is:
(a) y (b)
(c) L (d) a
4. A material of length L { at temperature Q x K
is subjected to a temperature rise of 6 K. The
coefficient of linear expansion of the ma¬
terial is a K -1 . The material expands by:
(a) L 2 (1 + aO)
(b) L x a{0-0 x )
(c) L x \\+a(0-6 { )]
(d) L { a6
5. Some iron has a coefficient of linear expan¬
sion of 12 x 10 -6 K 1 . A 100 mm length
of iron piping is heated through 20 K. The
pipe extends by:
(a) 0.24 mm (b) 0.024 mm
(c) 2.4 mm (d) 0.0024 mm
6. If the coefficient of linear expansion is A,
the coefficient of superficial expansion is B
and the coefficient of cubic expansion is C,
which of the following is false?
Part Four
Part Four
270 Mechanical Engineering Principles
(a) C=3A (b) A = 5/2
(c) 5 = - C (d) A = C/3
2
7. The length of a 100 mm bar of metal in¬
creases by 0.3 mm when subjected to a
temperature rise of 100 K. The coefficient
of linear expansion of the metal is:
(a) 3 x 1(T 3 Kr 1 (b) 3 x 1(H K 1
(c) 3 x 1(T 5 Kr 1 (d) 3 x KT 6 Kr 1
(b) Bimetallic strips are used in thermo¬
stats, a thermostat being a temperature-
operated switch.
(c) As the temperature of water is de¬
creased from 4°C to 0°C contraction
occurs.
(d) A change of temperature of 15°C is
equivalent to a change of temperature
of 15 K.
8. A liquid has a volume V x at temperature Q v
The temperature is increased to 0 2 . If y is
the coefficient of cubic expansion, the in¬
crease in volume is given by:
(a) V x y(O 2 -0 l )
(b) V x yd ,
(c) V x + V x y 0 2
(d) V l \l+y(d 2 -0 l )\
9. Which of the following statements is false?
(a) Gaps need to be left in lengths of rail¬
way lines to prevent buckling in hot
weather.
The volume of a rectangular block of iron
at a temperature t x is V v The temperature is
raised to t 2 and the volume increases to V 2 .
If the coefficient of linear expansion of iron
is a , then volume V l is given by:
(a) V 2 [l + a(t 2 -t l )]
V 2
(b) l + 3a(? 2 -?j)
(c) 3 V 2 a{t 2
1 + a(t 2
(d) — V "
V,
For fully worked solutions to each of the problems in Practice Exercises 122 to 125 in this chapter,
go to the website:
www.routledge.com/cw/bird
Revision Test 8 Heat energy and transfer, and thermal expansion
brackets at the end of each question.
1. A block of aluminium having a mass of 20 kg
cools from a temperature of 250°C to 80°C.
How much energy is lost by the aluminium?
Assume the specific heat capacity of aluminium is
950 J/(kg °C). (5)
2. Calculate the heat energy required to convert
completely 12 kg of water at 30°C to superheated
steam at 100°C. Assume that the specific heat
capacity of water is 4200 J/(kg °C), and the
specific latent heat of vaporisation of water is
2260 kJ/(kg °C). (7)
A copper overhead transmission line has a length
of 60 m between its supports at 15°C. Calculate
its length at 40°C, if the coefficient of linear
expansion of copper is 17 x 10~ 6 K -1 . (6)
4. A gold sphere has a diameter of 40 mm at a tem¬
perature of 285 K. If the temperature of the sphere
is raised to 785 K, determine the increase in
(a) the diameter (b) the surface area (c) the vol¬
ume of the sphere. Assume the coefficient of
linear expansion for gold is 14 x 10~ 6 K _1 . (12)
This Revision Test covers the material contained in Chapters 21 and 22. The marks for each question are shown in
3.
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 8,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
Part Four
Chapter 23
Why it is important to understand: Hydrostatics
This chapter describes fluid pressure, together with buoyancy and hydrostatic stability. The chapter also
defines Archimedes’principle, which is used to determine the buoyancy of boats, yachts, ships, etc. The chapter
also describes metacentric height, which is used to determine the hydrostatic stability of the aforementioned
vessels, and the explanation of this topic is aided with a number of simple worked examples. The chapters also
describe gauges used in fluid mechanics, such as barometers, manometers, and the Bourdon pressure and
vacuum gauges. These gauges are used to determine the properties and behaviour of fluids when they are met
in practice. Calculations are given of simple floating structures and reference is made to the mid-ordinate rule,
described earlier in Chapter 16, which can be used for determining the areas and volumes of complex shapes,
such as those often met in naval architecture and civil engineering and many other branches of engineering.
A knowledge of hydrostatics is of importance in boats, ships and other floating bodies and structures.
At the end of this chapter you should be able to:
• define pressure and state its unit
• understand pressure in fluids
• distinguish between atmospheric, absolute and gauge pressures
• state and apply Archimedes’ principle
• describe the construction and principle of operation of different types of barometer
• describe the construction and principle of operation of different types of manometer
• describe the construction and principle of operation of the Bourdon pressure gauge
• describe the construction and principle of operation of different types of vacuum gauge
• calculate hydrostatic pressure on submerged surfaces
• understand hydrostatic thrust on curved surfaces
• define buoyancy
• appreciate and perform calculations on the stability of floating bodies
23.1 Pressure
The pressure acting on a surface is defined as the per¬
pendicular force per unit area of surface. The unit of
pressure is the pascal*, Pa, where 1 pascal is equal to
1 newton per square metre. Thus pressure,
P = -j pascals
where F is the force in newtons acting at right angles to
a surface of area A square metres.
When a force of 20 N acts uniformly over, and per¬
pendicular to, an area of 4 m 2 , then the pressure on the
area,/), is given by:
20 N
4m 2
= 5 Pa
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
Hydrostatics 273
* ilaise Pascal (19 June 1623-19 August 1662) was a French
polymath. A child prodigy, he wrote a significant treatise on the
subject of projective geometry at the age of sixteen, and later
corresponded with Pierre de Fermat on probability theory, strongly
influencing the development of modem economics and social
science. To find out more go to www.routledge.com/cw/bird
It should be noted that for irregular shaped flat sur¬
faces, such as the water planes of ships, their areas can
be calculated using the mid-ordinate rule, described in
Chapter 16.
A table loaded with books has a force
of 250 N acting in each of its legs. If the contact
area between each leg and the floor is 50 mm 2 , find
the pressure each leg exerts on the floor.
From above, pressure p =
force
area
Hence,
250 N _ 250 N
50 mm 2 50xl0~ 6 m 2
= 5 x 10 6 N/m 2
= 5 MPa
That is, the pressure exerted by each leg on the floor
is 5 MPa
Calculate the force exerted by the atmo¬
sphere on a pool of water that is 30 m long by 10 m
wide, when the atmospheric pressure is 100 kPa.
force
From above, pressure =-
area
hence, force = pressure x area.
The area of the pool is 30 m x 10 m = 300 m 2
Thus, force on pool, F = pressure x area
= 100 kPa x 300 m 2 and
since 1 Pa = 1 N/m 2 ,
F= (100 x 10 3 ) -2L x 300 m 2
m 2
= 3 x 10 7 N = 30 x 10 6 N
= 30 MN
That is, the force on the pool of water is 30 MN
A circular piston exerts a pressure
of 80 kPa on a fluid, when the force applied to the
piston is 0.2 kN. Find the diameter of the piston.
From above, pressure =
hence, area =
force
area
force
pressure
Force in newtons = 0.2 kN
= 0.2 x 10 3 N = 200 N, and
pressure in pascals = 80 kPa = 80000 Pa
= 80000 N/m 2 .
force 200 N
Hence, area = -=--
pressure 80000 N/m 2
= 0.0025 m 2
Since the piston is circular, its area is given by 7rd 2 /4,
where d is the diameter of the piston.
nd 2
Hence, area = —-— = 0.0025
4
from which, d 2 = 0.0025 x-= 0.003183
71
i.e. d = Vo.003183 = 0.0564 m
= 56.4 mm
Hence, the diameter of the piston is 56.4 mm
Now try the following Practice Exercise
Practice Exercise 126 Further problems on
pressure
1. A force of 280 N is applied to a piston of
a hydraulic system of cross-sectional area
Part Four
Part Four
274 Mechanical Engineering Principles
O.OIO m 2 . Determine the pressure produced
by the piston in the hydraulic fluid.
[28 kPa]
2. Find the force on the piston of question 1 to
produce a pressure of 450 kPa. [4.5 kN]
3. If the area of the piston in question 1 is
halved and the force applied is 280 N, deter¬
mine the new pressure in the hydraulic fluid.
[56 kPa]
23.2 Fluid pressure
A fluid is either a liquid or a gas and there are four basic
factors governing the pressure within fluids.
Figure 23.1
(a) The pressure at a given depth in a fluid is equal in
all directions; see Figure 23.1(a).
(b) The pressure at a given depth in a fluid is inde¬
pendent of the shape of the container in which the
fluid is held. In Figure 23.1(b), the pressure at A is
the same as the pressure at Y.
(c) Pressure acts at right angles to the surface contain¬
ing the fluid. In Figure 23.1(c), the pressures at
points A to F all act at right angles to the container.
(d) When a pressure is applied to a fluid, this pressure
is transmitted equally in all directions. In Fig¬
ure 23.1(d), if the mass of the fluid is neglected,
the pressures at points AtoD are all the same.
The pressure,/?, at any point in a fluid depends on three
factors:
(a) the density of the fluid, /?, in kg/m 3
(b) the gravitational acceleration, g, taken as approxi¬
mately 9.8 m/s 2 (or the gravitational field force
in N/kg), and
(c) the height of fluid vertically above the point, h
metres.
The relationship connecting these quantities is:
p = pgh pascals
When the container shown in Figure 23.2 is filled with
water of density 1000 kg/m 3 , the pressure due to the
water at a depth of 0.03 m below the surface is given by:
p= pgh = (1000 x 9.8 x 0.03)Pa = 294 Pa
0.03 m
Figure 23.2
In the case of the Mariana Trench, which is situated in
the Pacific ocean, near Guam, the hydrostatic pressure
is about 115.2 MPa or 1152 bar, where 1 bar = 10 5 Pa
and the density of sea water being 1020 kg/m 3 .
A tank contains water to a depth of
600 mm. Calculate the water pressure (a) at a depth
of 350 mm, and (b) at the base of the tank. Take the
density of water as 1000 kg/m 3 and the gravita¬
tional acceleration as 9.8 m/s 2 .
From above, pressure p at any point in a fluid is given
by p = pgh pascals, where p is the density in kg/m 3 , g is
the gravitational acceleration in m/s 2 and h is the height
of fluid vertically above the point in metres.
(a) At a depth of 350 mm = 0.35 m,
p = pgh = 1000 x 9.8 x 0.35
= 3430 Pa = 3.43 kPa
(b) At the base of the tank, the vertical height of the
water is 600 mm = 0.6 m.
Hence, p = 1000 x 9.8 x 0.6
= 5880 Pa = 5.88 kPa
A storage tank contains petrol to a
height of 4.7 m. If the pressure at the base of the
tank is 32.2 kPa, determine the density of the pet¬
rol. Take the gravitational field force as 9.8 m/s 2 .
From above, pressure p = pgh pascals, where p is the
density in kg/m 3 , g is the gravitational acceleration in
m/s 2 and h is the vertical height of the petrol in metres.
Transposing gives:
P
gh
Pressure p is 32.2 kPa= 32200 Pa
32200
hence, density, p =-
’ H 9.8 x 4.7
That is, the density of the petrol is 699 kg/m 3 .
699 kg/nr
Hydrostatics 275
A vertical tube is partly filled with
mercury of density 13600 kg/m 3 4 . Find the height,
in millimetres, of the column of mercury, when the
pressure at the base of the tube is 101 kPa. Take the
gravitational field force as 9.8 m* 1 /s 2 .
From above, pressure p = pgh , hence vertical height h
is given by:
Pg
Pressure p = 101 kPa = 101000 Pa,
101000
thuS ’ h = 13600x 9.8 = °- 758 m
That is, the height of the column of mercury is 758 mm
Now try the following Practice Exercise
Practice Exercise 127 Further problems on
fluid pressure
(Take the gravitational acceleration as 9.8 m/s 2 )
1. Determine the pressure acting at the base of
a dam, when the surface of the water is 35 m
above base level. Take the density of water
as 1000 kg/m 3 . [343 kPa]
2. An uncorked bottle is full of sea water of
density 1030 kg/m 3 . Calculate, correct to
3 significant figures, the pressures on the
side wall of the bottle at depths of (a) 30 mm,
and (b) 70 mm below the top of the bottle.
[(a) 303 Pa (b) 707 Pa]
3. A £/-tube manometer is used to determine
the pressure at a depth of 500 mm below
the free surface of a fluid. If the pressure
at this depth is 6.86 kPa, calculate the den¬
sity of the liquid used in the manometer.
[1400 kg/m 3 ]
4. A submarine pressure hull in the form of a
circular cylinder is of external diameter 10 m
and length 200 m. It dives to the bottom of
the Mariana Trench which is 11.52 km deep.
What will be the mass of water acting on the
submarine’s circular surface in terms of the
number of London double-decker buses, giv¬
en that the mass of a London double-decker
bus is 7 tonnes. Assume that the density of
water, p = 1020 kg/m 3 and gravitational ac¬
celeration, g = 9.81 m/s 2 . [10.55 million]
23.3 Atmospheric pressure
The air above the Earth’s surface is a fluid, having a
density,/?, which varies from approximately 1.225 kg/m 3
at sea level to zero in outer space. Since p= pgh , where
height h is several thousands of metres, the air exerts
a pressure on all points on the Earth’s surface. This
pressure, called atmospheric pressure, has a value of
approximately 100 kilopascals (or 1 bar). Two terms
are commonly used when measuring pressures:
(a) absolute pressure, meaning the pressure above
that of an absolute vacuum (i.e. zero pressure), and
(b) gauge pressure, meaning the pressure above that
normally present due to the atmosphere.
Thus, absolute pressure = atmospheric pressure +
gauge pressure.
Thus, a gauge pressure of 50 kPa is equivalent to an
absolute pressure of (100 + 50) kPa, i.e. 150 kPa, since
the atmospheric pressure is approximately 100 kPa.
Calculate the absolute pressure at a
point on a submarine, at a depth of 30 m below the
surface of the sea, when the atmospheric pressure
is 101 kPa. Take the density of sea water as
1030 kg/m 3 and the gravitational acceleration as
9.8 m/s 2 .
From Section 23.2, the pressure due to the sea, that is,
the gauge pressure (p ) is given by:
o
Pa = Pgh Pascals
o
i.e. p a = 1030 x 9.8 x 30 = 302820 Pa = 302.82 kPa
o
From above, absolute pressure
= atmospheric pressure + gauge pressure
= (101 + 302.82) kPa - 403.82 kPa
That is, the absolute pressure at a depth of 30 m is
403.82 kPa
Part Four
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276 Mechanical Engineering Principles
Now try the following Practice Exercise
Practice Exercise 128 Further problems
on atmospheric
pressure
Take the gravitational acceleration as 9.8 m/s 2 ,
the density of water as 1000 kg/m 3 , and the den¬
sity of mercury as 13600 kg/m 3 .
1. The height of a column of mercury in a
barometer is 750 mm. Determine the at¬
mospheric pressure, correct to 3 significant
figures. [100 kPa]
2. A (7-tube manometer containing mercury
gives a height reading of 250 mm of mer¬
cury when connected to a gas cylinder. If
the barometer reading at the same time is
756 mm of mercury, calculate the absolute
pressure of the gas in the cylinder, correct to
3 significant figures. [134 kPa]
3. A water manometer connected to a condens¬
er shows that the pressure in the condenser
is 350 mm below atmospheric pressure. If
the barometer is reading 760 mm of mer¬
cury, determine the absolute pressure in the
condenser, correct to 3 significant figures.
[97.9 kPa]
If a body floats on the surface of a liquid all of its
weight appears to have been lost. The weight of liquid
displaced is equal to the weight of the floating body.
A body weighs 2.760 N in air and
1.925 N when completely immersed in water of
density 1000 kg/m 3 . Calculate (a) the volume of
the body (b) the density of the body and (c) the
relative density of the body. Take the gravitational
acceleration as 9.81 m/s 2 .
(a) The apparent loss of weight is 2.760 N - 1.925 N
= 0.835 N. This is the weight of water displaced,
i.e. Vpg , where V is the volume of the body and
p is the density of water,
i.e. 0.835 N= Vx 1000 kg/m 3 x 9.81 m/s 2
- V x 9.81 kN/m 3
0.835 ,
Hence, V= --m
9.81 x 10 3
= 8.512 x 10" 5 m 3
- 8.512 x 10 4 mm 3
(b) The density of the body
_ mass _ weight
volume gxf
__ 2.760N _
9.81m/s 2 x 8.512 xl0" 5 m 3
4. A Bourdon pressure gauge shows a pressure
of 1.151 MPa. If the absolute pressure is
1.25 MPa, find the atmospheric pressure in
millimetres of mercury. [743 mm]
23.4 Archimedes'principle
Archimedes’ principle* states that:
If a solid body floats , or is submerged , in a liquid ,
the liquid exerts an upthrust on the body equal to the
gravitational force on the liquid displaced by the body.
In other words, if a solid body is immersed in a liquid,
the apparent loss of weight is equal to the weight of
liquid displaced.
If V is the volume of the body below the surface
of the liquid, then the apparent loss of weight W is
given by:
W= Vco = Vpg
where co is the specific weight (i.e. weight per unit
volume) and p is the density.
2.760
9.81
kg x 10 ;
8.512m
3305 kg/nr
= 3.305 tonne/m 3
density
Relative density = --:---
density ot water
* Archimedes of Syracuse (c. 287 BC-c. 212 BC) was a Greek
mathematician, physicist, engineer, inventor and astronomer,
credited with designing innovative machines, including siege
engines. To find out more go to www.routledge.com/cw/bird
Hydrostatics 277
Hence, the relative density of the body
3305 kg/m 3
1000 kg/m
= 3.305
A rectangular watertight box is
560 mm long, 420 mm wide and 210 mm deep.
It weighs 223 N. (a) If it floats with its sides and
ends vertical in water of density 1030 kg/m 3 , what
depth of the box will be submerged? (b) If the box
is held completely submerged in water of density
1030 kg/m 3 , by a vertical chain attached to the un¬
derside of the box, what is the force in the chain?
(a) The apparent weight of a floating body is zero.
That is, the weight of the body is equal to the
weight of liquid displaced. This is given by:
Vpg
where V is the volume of liquid displaced, and p
is the density of the liquid.
Here, 223 N = Vx 1030 kg/m 3 x 9.81 m/s 2
= Vx 10.104 kN/m 3
223 N
Hence, V= -t
10.104 kN/m 3
= 22.07 x 10~ 3 m 3
This volume is also given by Lbd , where
L = length of box, b = breadth of box, and
d = depth of box submerged,
i.e. 22.07 x 10 -3 m 3 = Lx b x d
= 0.56 m x 0.42 m x d
Hence, depth submerged, d = 22.07 x 10—
0.56x0.42
= 0.09384 m -93.84 mm
(b) The volume of water displaced is the total vol¬
ume of the box. The upthrust or buoyancy of the
water, i.e. the ‘apparent loss of weight’, is greater
than the weight of the box. The force in the chain
accounts for the difference.
Volume of water displaced,
V— 0.56 m x 0.42 m x 0.21 m
= 4.9392 x 10 -2 m 3
Weight of water displaced
= Vpg = 4.9392 x 10" 2 m 3 x 1030 kg/m 3 x 9.81 m/s 2
= 499.1 N
The force in the chain
= weight of water displaced - weight of box
= 499.1 N - 223 N = 276.1 N
Now try the following Practice Exercise
Practice Exercise 129 Further problems
on Archimedes'
principle
Take the gravitational acceleration as 9.8 m/s 2 ,
the density of water as 1000 kg/m 3 and the den¬
sity of mercury as 13600 kg/m3.
1. A body of volume 0.124 m 3 is completely
immersed in water of density 1000 kg/m 3 .
What is the apparent loss of weight of the
body? [1.215 kN]
2. A body of weight 27.4 N and volume
1240 cm 2 is completely immersed in water
of specific weight 9.81 kN/m 3 . What is its
apparent weight? [ 15.25 N]
3. A body weighs 512.6 N in air and 256.8 N
when completely immersed in oil of density
810 kg/m 3 . What is the volume of the body?
[0.03222 m 3 ]
4. A body weighs 243 N in air and 125 N when
completely immersed in water. What will it
weigh when completely immersed in oil of
relative density 0.8? [148.6 N]
5. A watertight rectangular box, 1.2 m long and
0.75 m wide, floats with its sides and ends
vertical in water of density 1000 kg/m 3 . If
the depth of the box in the water is 280 mm,
what is its weight? [2.47 kN]
6. A body weighs 18 N in air and 13.7 N when
completely immersed in water of density
1000 kg/m 3 . What is the density and relative
density of the body?
[4.186 tonne/m 3 , 4.186]
7. A watertight rectangular box is 660 mm long
and 320 mm wide. Its weight is 336 N. If it
floats with its sides and ends vertical in water
of density 1020 kg/m 3 , what will be its depth
in the water? [159 mm]
8. A watertight drum has a volume of 0.165 m 3
and a weight of 115 N. It is completely
submerged in water of density 1030 kg/m 3 ,
held in position by a single vertical chain
attached to the underside of the drum. What
is the force in the chain? [1.551 kN]
Part Four
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278 Mechanical Engineering Principles
23.5 Measurement of pressure
As stated earlier, pressure is the force exerted by a fluid
per unit area. A fluid (i.e. liquid, vapour or gas) has a
negligible resistance to a shear force, so that the force
it exerts always acts at right angles to its containing
surface.
The SI unit of pressure is the pascal, Pa, which is unit
force per unit area, i.e. 1 Pa = 1 N/m 2
The pascal is a very small unit and a commonly used
larger unit is the bar, where 1 bar = 10 5 Pa
Atmospheric pressure is due to the mass of the air
above the Earth’s surface being attracted by Earth’s
gravity. Atmospheric pressure changes continuously.
A standard value of atmospheric pressure, called
‘standard atmospheric pressure’, is often used, having
a value of 101325 Pa or 1.01325 bars or 1013.25 milli¬
bars. This latter unit, the millibar, is usually used in the
measurement of meteorological pressures. (Note that
when atmospheric pressure varies from 101325 Pa it is
no longer standard.)
Pressure indicating instruments are made in a wide
variety of forms because of their many different
applications. Apart from the obvious criteria such
as pressure range, accuracy and response, many mea¬
surements also require special attention to material,
sealing and temperature effects. The fluid whose pres¬
sure is being measured may be corrosive or may be at
high temperatures. Pressure indicating devices used in
science and industry include:
(i) barometers (see Section 23.6),
(ii) manometers (see Section 23.8),
(iii) Bourdon pressure gauge (see Section 23.9), and
(iv) McLeod and Pirani gauges (see Section 23.10).
23.6 Barometers
Introduction
A barometer is an instrument for measuring atmo¬
spheric pressure. It is affected by seasonal changes of
temperature. Barometers are therefore also used for the
measurement of altitude and also as one of the aids in
weather forecasting. The value of atmospheric pres¬
sure will thus vary with climatic conditions, although
not usually by more than about 10% of standard atmo¬
spheric pressure.
Construction and principle of operation
A simple barometer consists of a glass tube, just less
than 1 m in length, sealed at one end, filled with mer¬
cury and then inverted into a trough containing more
mercury. Care must be taken to ensure that no air enters
the tube during this latter process. Such a barometer
is shown in Figure 23.3(a) and it is seen that the level
of the mercury column falls, leaving an empty space,
called a vacuum. Atmospheric pressure acts on the
surface of the mercury in the trough as shown and this
pressure is equal to the pressure at the base of the col¬
umn of mercury in the inverted tube, i.e. the pressure of
the atmosphere is supporting the column of mercury.
If the atmospheric pressure falls the barometer height
h decreases. Similarly, if the atmospheric pressure
rises, then h increases. Thus atmospheric pressure can
be measured in terms of the height of the mercury
column. It may be shown that for mercury the height
h is 760 mm at standard atmospheric pressure, i.e.
Atmospheric
pressure
I t
Vacuum
T
Mercury
Barometric
height, h
Trough
(a)
Figure 23.3
Hydrostatics 279
a vertical column of mercury 760 mm high exerts a
pressure equal to the standard value of atmospheric
pressure.
There are thus several ways in which atmospheric pres¬
sure can be expressed:
Standard atmospheric pressure
= 101325 Pa or 101.325 kPa
= 101325 N/m 2 or 101.325 kN/m 2
= 1.01325 bars or 1013.25 mbars
= 760 mm of mercury
Another arrangement of a typical barometer is shown
in Figure 23.3(b) where a U -tube is used instead
of an inverted tube and trough, the principle being
similar.
If, instead of mercury, water was used as the liquid
in a barometer, then the barometric height h at standard
atmospheric pressure would be 13.6 times more than
for mercury, i.e. about 10.4 m high, which is not very
practicable. This is because the relative density of mer¬
cury is 13.6.
Types of barometer
The Fortin* barometer is an example of a mercury
barometer that enables barometric heights to be mea¬
sured to a high degree of accuracy (in the order of one-
tenth of a millimetre or less). Its construction is merely
a more sophisticated arrangement of the inverted tube
and trough shown in Figure 23.3(a), with the addition
of a vernier scale to measure the barometric height with
great accuracy. A disadvantage of this type of barom¬
eter is that it is not portable.
A Fortin barometer is shown in Figure 23.4.
Mercury is contained in a leather bag at the base of the
mercury reservoir, and height, //, of the mercury in the
reservoir can be adjusted using the screw at the base of
the barometer to depress or release the leather bag. To
measure the atmospheric pressure the screw is adjusted
until the pointer at H is just touching the surface of
the mercury and the height of the mercury column is
then read using the main and vernier scales. The
measurement of atmospheric pressure using a Fortin
barometer is achieved much more accurately than by
using a simple barometer.
* lean Nicolas Fortin (1750-1831) was a French maker of
scientific instruments and is chiefly remembered for his design
of the barometer. To find out more go to www.routledge.com/
cw/bird
Main
scale
Vernier
scale
Barometric
height in
millimetres
of mercury
H
Mercury
reservoir
Figure 23.4
A portable type often used is the aneroid barometer. Such
a barometer consists basically of a circular, hollow, sealed
vessel, S , usually made from thin flexible metal. The air
pressure in the vessel is reduced to nearly zero before
sealing, so that a change in atmospheric pressure will
cause the shape of the vessel to expand or contract. These
small changes can be magnified by means of a lever and
be made to move a pointer over a calibrated scale. Figure
23.5 shows a typical arrangement of an aneroid barometer.
The scale is usually circular and calibrated in millimetres
of mercury. These instruments require frequent calibration.
Figure 23.5
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280 Mechanical Engineering Principles
23.7 Absolute and gauge pressure
A barometer measures the true or absolute pressure
of the atmosphere. The term absolute pressure means
the pressure above that of an absolute vacuum (which
is zero pressure), as stated earlier. In Figure 23.6 a
pressure scale is shown with the line AB represent¬
ing absolute zero pressure (i.e. a vacuum) and line CD
representing atmospheric pressure. With most practical
pressure-measuring instruments the part of the instru¬
ment that is subjected to the pressure being measured is
also subjected to atmospheric pressure. Thus practical
instruments actually determine the difference between
the pressure being measured and atmospheric pres¬
sure. The pressure that the instrument is measuring is
then termed the gauge pressure. In Figure 23.6, the line
EF represents an absolute pressure which has a value
greater than atmospheric pressure, i.e. the ‘gauge’ pres¬
sure is positive.
Pressure
scale
^ | Positive gauge F
pressure
Atmospheric pressure
-*-n
Negative gauge u
Absolute
pressure
®f
Absolute
pressure
i
pressure
H
Absolute zero pressure
(complete vacuum)
B
Figure 23.6
Thus, absolute pressure = gauge pressure +
atmospheric pressure.
Hence a gauge pressure of, say, 60 kPa recorded on an
indicating instrument when the atmospheric pressure
is 101 kPa is equivalent to an absolute pressure of
60 kPa + 101 kPa, or 161 kPa.
Pressure-measuring indicating instruments are re¬
ferred to generally as pressure gauges (which acts as a
reminder that they measure ‘gauge’ pressure).
It is possible, of course, for the pressure indicated on
a pressure gauge to be below atmospheric pressure, i.e.
the gauge pressure is negative. Such a gauge pressure
is often referred to as a vacuum, even though it does
not necessarily represent a complete vacuum at abso¬
lute zero pressure. Such a pressure is shown by the line
GH in Figure 23.6. An indicating instrument used for
measuring such pressures is called a vacuum gauge.
A vacuum gauge indication of, say, 0.4 bar, means
that the pressure is 0.4 bar less than atmospheric pres¬
sure. If atmospheric pressure is 1 bar, then the absolute
pressure is 1 - 0.4 or 0.6 bar.
23.8 The manometer
A manometer is a device for measuring or comparing
fluid pressures, and is the simplest method of indicat¬
ing such pressures.
U -tube manometer
A (7-tube manometer consists of a glass tube bent into
a U shape and containing a liquid such as mercury.
A (7-tube manometer is shown in Figure 23.7(a). If
limb A is connected to a container of gas whose pres¬
sure is above atmospheric, then the pressure of the gas
will cause the levels of mercury to move as shown in
Figure 23.7(b), such that the difference in height is
hy The measuring scale can be calibrated to give the
gauge pressure of the gas as h { mm of mercury.
Measuring
scale
Figure 23.7
Hydrostatics 281
If limb A is connected to a container of gas whose
pressure is below atmospheric then the levels of mer¬
cury will move as shown in Figure 23.7(c), such that
their pressure difference is h 2 mm of mercury.
It is also possible merely to compare two pres¬
sures, say, P 4 and P B , using a U -tube manometer.
Figure 23.7(d) shows such an arrangement with
{P B - P A ) equivalent to h mm of mercury. One appli¬
cation of this differential pressure-measuring device
is in determining the velocity of fluid flow in pipes
(see Chapter 24).
For the measurement of lower pressures, water
or paraffin may be used instead of mercury in the
£/-tube to give larger values of h and thus greater
sensitivity.
Inclined manometers
For the measurement of very low pressures, greater
sensitivity is achieved by using an inclined manom¬
eter, a typical arrangement of which is shown in Fig¬
ure 23.8. With the inclined manometer the liquid used
is water and the scale attached to the inclined tube is
calibrated in terms of the vertical height h. Thus when
a vessel containing gas under pressure is connected
to the reservoir, movement of the liquid levels of the
manometer occurs. Since small-bore tubing is used the
movement of the liquid in the reservoir is very small
compared with the movement in the inclined tube and
is thus neglected. Hence the scale on the manometer is
usually used in the range 0.2 mbar to 2 mbar.
Inclined manometer
Figure 23.8
The pressure of a gas that a manometer is capable
of measuring is naturally limited by the length of
tube used. Most manometer tubes are less than 2 m
in length and this restricts measurement to a maxi¬
mum pressure of about 2.5 bar (or 250 kPa) when
mercury is used.
23.9 The Bourdon pressure gauge
Pressures many times greater than atmospheric can
be measured by the Bourdon pressure gauge, which
is the most extensively used of all pressure-indicating
instruments. It is a robust instrument. Its main
component is a piece of metal tube (called the Bourdon
tube), usually made of phosphor bronze or alloy steel,
of oval or elliptical cross-section, sealed at one end and
bent into an arc. In some forms the tube is bent into a
spiral for greater sensitivity. A typical arrangement is
shown in Figure 23.9(a). One end, E , of the Bourdon
tube is fixed and the fluid whose pressure is to be
measured is connected to this end. The pressure acts
at right angles to the metal tube wall as shown in the
cross-section of the tube in Figure 23.9(b). Because of
its elliptical shape it is clear that the sum of the pressure
components, i.e. the total force acting on the sides A and
C, exceeds the sum of the pressure components acting
on ends B and D. The result is that sides A and C tend
to move outwards and B and D inwards tending to form
Graduated
scale
Toothed
segment
Bourdon
tube
Fluid pressure
to be measured
(a)
A
B
Figure 23.9
Part Four
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282 Mechanical Engineering Principles
a circular cross-section. As the pressure in the tube is
increased the tube tends to uncurl, or if the pressure
is reduced the tube curls up further. The movement
of the free end of the tube is, for practical purposes,
proportional to the pressure applied to the tube, this
pressure, of course, being the gauge pressure (i.e. the
difference between atmospheric pressure acting on the
outside of the tube and the applied pressure acting on
the inside of the tube). By using a link, a pivot and
a toothed segment as shown in Figure 23.9(a), the
movement can be converted into the rotation of a
pointer over a graduated calibrated scale.
The Bourdon tube pressure gauge is capable of
measuring high pressures up to 10 4 bar (i.e. 7600 m of
mercury) with the addition of special safety features.
A pressure gauge must be calibrated, and this is done
either by a manometer, for low pressures, or by a piece
of equipment called a ‘dead weight tester’. This tester
consists of a piston operating in an oil-filled cylinder
of known bore, and carrying accurately known weights
as shown in Figure 23.10. The gauge under test is at¬
tached to the tester and a screwed piston or ram applies
the required pressure, until the weights are just lifted.
While the gauge is being read, the weights are turned to
reduce friction effects.
gauges. The basic principle of this gauge is that it
takes a known volume of gas at a pressure so low that
it cannot be measured, then compresses the gas in a
known ratio until the pressure becomes large enough to
be measured by an ordinary manometer. This device is
used to measure low pressures, often in the range 10 -6 to
1.0 mm of mercury. A disadvantage of the McLeod
gauge is that it does not give a continuous reading
of pressure and is not suitable for registering rapid
variations in pressure.
Pirani gauge
The Pirani* gauge measures the resistance and thus
the temperature of a wire through which current is
flowing. The thermal conductivity decreases with the
pressure in the range 10 _1 to 10 -4 mm of mercury so
that the increase in resistance can be used to measure
pressure in this region. The Pirani gauge is calibrated
by comparison with a McLeod gauge.
23.11 Hydrostatic pressure on
submerged surfaces
From Section 23.2, it can be seen that hydrostatic pres¬
sure increases with depth according to the formula:
p =pg>t
The deepest part of the ocean is the
Mariana Trench, where its depth is approximately
11.52 km (7.16 miles). What is the gauge pressure
at this depth, assuming that p = 1020 kg/m 3 and
g = 9.81 m/s 2 ?
Gauge pressure, p = pgh
Figure 23.10
23.10 Vacuum gauges
Vacuum gauges are instruments for giving a visual in¬
dication, by means of a pointer, of the amount by which
the pressure of a fluid applied to the gauge is less than
the pressure of the surrounding atmosphere. Two ex¬
amples of vacuum gauges are the McLeod gauge and
the Pirani gauge.
McLeod gauge
The McLeod* gauge is normally regarded as a
standard and is used to calibrate other forms of vacuum
= 1020^- x 9.81 — x 11.52 x 10 3 m
m 3 s 2
n 0 1 bar
= 11.527 x 10 7 N/m 2 x —^ T
10 5 N/m 2
i.e. pressure,/; = 1152.7 bar
* erbert McLeod (February 1841-October 1923) was a Brit¬
ish chemist, noted for the invention of the McLeod gauge and
for the invention of the sunshine recorder. To find out more go
to www.routledge.com/cw/bird
*Marcello Stefano Pirani (1 July 1880-11 January 1968) was
a German physicist known for his invention of the Pirani vacu¬
um gauge. To find out more go to www.routledge.com/cw/bird
Hydrostatics 283
Note that from the above calculation, it can be seen that
a gauge pressure of 1 bar is approximately equivalent
to a depth of 10 m.
Determine an expression for the
thrust acting on a submerged plane surface, which
is inclined to the horizontal by an angle 0 , as
shown in Figure 23.11.
From Figure 23.11, SF = elemental thrust on dA
= pgh x dA
But h=y sin 0
Hence, SF = pgy sin 6 dA
Total thrust on plane surface = F= \ dF= \ pgy sin 0 dA
or
F = pg sin
However, J y dA =Ah where A = area of the surface,
and h = distance of the centroid of the plane from the
free surface.
Determine an expression for the
position of the centre of pressure of the plane
surface P(x', y') of Figure 23.11; this is also
the position of the centre of thrust.
Taking moments about O gives:
F y = JpgysinOdA x y
However, F = pg sin 0 J ydA
Hence,
f pg)’ * 1 2 sin0dA pgsinO Jy 2 dA
pg sin 0 J ydA pg sin 0 J ydA
(Ak 2 )
Ox
Ay
where
Ak
Ox
= the second moment of area
about Ox
k = the radius of gyration from O.
Now try the following Practice Exercise
Practice Exercise 130 Further problems on
hydrostatic pressure
on submerged
surfaces
(Takeg = 9.81 m/s 2 )
1. Determine the gauge pressure acting on the
surface of a submarine that dives to a depth
of 500 m. Take water density as 1020 kg/m 3 4 5 6 .
[50.03 bar]
2. Solve Problem 1, when the submarine dives
to a depth of 780 m. [78.05 bar]
3. If the gauge pressure measured on the sur¬
face of the submarine of Problem 1 were
92 bar, at what depth has the submarine
dived to? [919.4 m]
4. A tank has a flat rectangular end, which is of
size 4 m depth by 3 m width. If the tank is
filled with water to its brim and the flat end is
vertical, determine the position of its centre
of pressure and the thrust on this end. Take
water density as 1000 kg/m 3 .
[2.667 m ; 0.235 MN]
5. If another vertical flat rectangular end of the
tank of Problem 4 is of size 6 m depth by
4 m width, determine the position of its cen¬
tre of pressure and the thrust on this end. The
depth of water at this end may be assumed to
be 6 m. [4 m ; 0.706 MN]
6. A tank has a flat rectangular end, which is
inclined to the horizontal surface, so that
Part Four
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284 Mechanical Engineering Principles
6 = 30°, where 6 is as defined in Figure
23.11, page 283. If this end is of size 6 m
height and 4 m width, determine the position
of its centre of pressure from the top and the
thrust on this end. The tank may be assumed
to be just full.
[2 m ; 0.235 MN]
23.12 Hydrostatic thrust on
curved surfaces
As hydrostatic pressure acts perpendicularly to a sur¬
face, the integration of SF over the surface can be
complicated. One method of determining the thrust on
a curved surface is to project its area on flat vertical
and horizontal surfaces, as shown by AB and DE ,
respectively, in Figure 23.12.
Free surface D E
Figure 23.12
From equilibrium considerations, F = F x and W = F y
and these thrusts must act through the centre of pres¬
sures of the respective vertical and horizontal planes.
The resultant thrust can be obtained by adding F v and
F y vectorially, where W = weight of the fluid enclosed
by the curved surface and the vertical projection lines
to the free surface, and G = centre of gravity of W.
23.13 Buoyancy
The upward force exerted by the fluid on a body that
is wholly or partially immersed in it is called the buoy¬
ancy of the body.
23.14 The stability of floating bodies
For most ships and boats the centre of buoyancy ( B ) of
the vessel is usually below the vessels’ centre of grav¬
ity (G), as shown in Figure 23.13(a). When this vessel
is subjected to a small angle of keel ( 6 ), as shown in
Figure 23.13(b), the centre of buoyancy moves to the
position B\
where BM = the centre of curvature of the centre of
buoyancy = y , (given without proof)
GM = the metacentric height,
(a) (b)
Figure 23.13
M = the position of the metacentre,
/ = the second moment of area of the water plane
about its centreline, and
V= displaced volume of the vessel.
The metacentric height GM can be found by a simple
inclining experiment, where a weight P is moved trans¬
versely a distance x, as shown in Figure 23.14.
(a) (b)
Figure 23.14
From rotational equilibrium considerations,
W(GM )tan 6 = Px
Hydrostatics 285
Therefore,
Px
GM= — cot 0
W
(23.1)
where W= the weight of the vessel,
1
and cot 0 = -—
tan#
A naval architect has carried out
hydrostatic calculations on a yacht, where he has
found the following:
M = mass of yacht =100 tonnes,
KB = vertical distance of the centre of buoyancy
(B) above the keel (K) = 1.2 m (see Figure
23.15),
BM = distance of the metacentre (M) above the
centre of buoyancy = 2.4 m.
Figure 23.15
He then carries out an inclining experiment, where
he moves a mass of 50 kg through a transverse
distance of 10 m across the yacht’s deck. In doing
this, he finds that the resulting angle of keel, 0=1°.
What is the metacentric height ( GM) and the posi¬
tion of the centre of gravity of the yacht above the
keel, namely KG? Assume g = 9.81 m/s 2 .
P = 50 kg x 9.81 =490.5 N,
W= 100 tonnes x 1000-^x9.81 — = 981 kN,
tonne s 2
x = 10 m,
0= 1° from which,
1
tan # = 0.017455 and cot # = --=57.29
tan#
From equation (23.1), GM =
Px
cot #
W
490.5NX 10m x 57.29
981x10 3 N
i.e. metacentric height, GM= 0.286 m
Now KM = KB + BM = 1.2 m + 2.4 m = 3.6 m
KG = KM- GM= 3.6 - 0.286 = 3.314 m
i.e. centre of gravity above the keel, KG = 3.314 m,
(where ‘ K ’ is a point on the keel).
A barge of length 30 m and width
8 m floats on an even keel at a depth of 3 m. What is
the value of its buoyancy? Take density of water, p ,
as 1000 kg/m 3 and g as 9.81 m/s 2 .
The displaced volume of the barge,
V= 30 m x 8 m x 3 m = 720 m 3 .
From Section 23.4,
buoyancy = Vpg = 720 m 3 x 1000-^- x 9.81
m 5 s z
= 7.063 MN
If the vertical centre of gravity of
the barge in Problem 14 is 2 m above the keel, (i.e.
KG = 2 m), what is the metacentric height of the
barge?
Now KB = the distance of the centre of buoyancy of the
3 m
barge from the keel = i.e. KB =1.5 m.
I Lb 3
From page 284, BM= — and for a rectangle, /= -jj-
from Table 9.1, page 133, where L = length of the wa-
terplane = 30 m, and
b = width of the waterplane = 8 m.
30 x 8 3 ,
Hence, moment of inertia, /= ——— = 1280 m
From Problem 14, volume, V— 720 m 3 ,
, / 1280 ,
hence, BM = — =- = 1.778 m
V 720
Now, KM = KB + BM= 1.5 m+ 1.778 m = 3.278 m
i.e. the metacentre above the keel, KM= 3.278 m.
Since KG = 2 m (given), then
GM = KM-KG = 3218-2 = 1.278 m,
i.e. the metacentric height of the barge, GM = 1.278 m
A circular cylindrical steel buoy,
made from 10 mm thick steel plate, is of a hollow
box-like disc shape, as shown in Figure 23.16. It is
sealed off at its top and bottom by circular plates so
that it is watertight, (a) If the external diameter of
the buoy, Z), is 1 m and its height, h, is 0.5 m, deter¬
mine its weight, W, given that the density of steel,
Part Four
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286 Mechanical Engineering Principles
p s = 7860 kg/m 3 , (b) At what depth, //, will the buoy
float if the density of water, p w — 1020 kg/m 3 ?
(c) What is its GM1 Take g to be 9.81 m/s 2 .
circular plate
Figure 23.16 Circular cylindrical buoy
(a) Weight of two end plates,
W j = kR 2 xtx p s x gx 2 where radius R = D/2
= k(0.5) 2 x (10 x 10" 3 ) x 7860 x 9.81 x 2
= 1211.2 N
Weight of circular cylinder,
W 2 = 27ri? mea n txhxp s xg
= 2nx (0.5 - 5 x 10“ 3 ) x (10 x 1(T 3 )
x 0.5x7860x9.81 = 1199 N
Total weight of buoy, W= W x + W 2
= 1211 + 1199 = 2410 N
(b) Buoyancy, B = W = 24\0 'H = weight of water
displaced (1)
Let H= draught of water of the buoy, so that:
B = kR 2 H x p w x 9.81
= /r(0.5) 2 x H x 1020 x 9.81 = 7858.9 H (2)
Equating (1) and (2) gives: 7858.9 H = 2410
from which, depth, H =
2410
7858.9
= 0.307 m
H 0.307
KB =
— = -
—
2
2
I
kR 4
BM
X
V
4
R 2
0.5
1
kR 2 H
- = 0.204 m
4 H 4 x 0.307
KM = KB + BM= 0.153 + 0.204
= 0.357 m
GM = KM-KG = 0.357 -
0.5
~2
Hence, GM = 0.107 m
A submarine pressure hull is in
the form of a watertight circular cylindrical shell,
of length 80 m, blocked off by flat ends, and of
external diameter 10 m, and it descends to the
bottom of the Mariana Trench, which is 11.52 km
deep. What will be the hydrostatic pressure acting
on it? If a double-decker London bus is of mass
7 tonnes what will be the equivalent number of
double-decker London buses acting on this hull?
Density of water, p w = 1020 kg/m 3 . Take g to be
9.81 m/s 2 .
Hydrostatic pressure,
p = pgh = 1020x9.81 x 11520
115.27 xlO 6
= 115.27 MPa -
10 5 Pa/bar
= 1152.7 bar = 1152.7 atmosphere
(since 10 5 pascals = 1 bar = 14.5 psi)
Let A = area of the external surface of the pressure hull
assuming flat ends = nR 2 x 2 + 2k RL
10
— and
where R = external cylinder radius
L = length between its ends = 80 m
Hence, area A = n{5) 2 x 2 + 2k x 5 x 80
- 157.1 + 2513.3 = 2670.4 m 2
Total hydrostatic head on the submarine hull
=pxA = 115.27 x 10 6 x 2670.4 = 307817 MN
10 6 N kg
Hence, W= 307817 MN x-x— *—
MN 9.81N
= 3.1378 xl0 10 kgx
tonne
1000 kg
= 31.378 x 10 6 tonnes
Number of double-decker London buses,
31.378 xlO 6 ,
N= - - - = 4.48 x 10 6
Thus, N = 4.48 million equivalent weight of double-
decker London buses
Because of the huge heads suffered by submarine pres¬
sure hulls, they are one of the most difficult structures
to design (see [1], page 289).
Hydrostatics 287
Now try the following Practice Exercises
Practice Exercise 131 Further problems on
hydrostatics
(In the following problems, where necessary, take
g = 9.81 m/s 2 and density of water p = 1020 kg/m 3 )
1. A ship is of mass 10000 kg. If the ship floats
in the water, what is the value of its buoy¬
ancy? [98.1 kN]
2. A submarine may be assumed to be in the
form of a circular cylinder of 10 m external
diameter and of length 100 m. If the sub¬
marine floats just below the surface of the
water, what is the value of its buoyancy?
[78.59 MN]
3. A barge of length 20 m and of width 5 m
floats on an even keel at a depth of 2 m. What
is the value of its buoyancy? [2 MN]
4. An inclining experiment is carried out on the
barge of Problem 3 where a mass of 20 kg is
moved transversely across the deck by a dis¬
tance of 2.2 m. The resulting angle of keel is
0.8°. Determine the metacentric height, GM.
[0.0155 m]
5. Determine the value of the radius of curva¬
ture of the centre of buoyancy, namely, BM ,
for the barge of Problems 3 and 4, and hence
the position of the centre of gravity above
the keel, KG.
[BM= 1.0417 m, KG = 2.026 m]
6. If the submarine of Problem 2 floats so
that its top is 2 m above the water, determine
the centre of curvature of the centre of buoy¬
ancy, BM. [0.633 m]
Practice Exercise 132 Short-answer
questions on
hydrostatics
1. Define pressure.
2. State the unit of pressure.
3. Define a fluid.
4. State the four basic factors governing the
pressure in fluids.
5. Write down a formula for determining the
pressure at any point in a fluid in symbols,
defining each of the symbols and giving
their units.
6. What is meant by atmospheric pressure?
7. State the approximate value of atmospheric
pressure.
8. State what is meant by gauge pressure.
9. State what is meant by absolute pressure.
10. State the relationship between absolute,
gauge and atmospheric pressures.
11. State Archimedes’ principle.
12. Name four pressure measuring devices.
13. Standard atmospheric pressure is 101325 Pa.
State this pressure in millibars.
14. Briefly describe how a barometer operates.
15. State the advantage of a Fortin barometer
over a simple barometer.
16. What is the main disadvantage of a Fortin
barometer?
17. Briefly describe an aneroid barometer.
18. What is a vacuum gauge?
19. Briefly describe the principle of operation
of a C/-tube manometer.
20. When would an inclined manometer be
used in preference to a U -tube manometer?
21. Briefly describe the principle of operation
of a Bourdon pressure gauge.
22. What is a ‘dead weight tester’?
23. What is a Pirani gauge?
24. What is a McLeod gauge used for?
25. What is buoyancy?
26. What does the abbreviation BM mean?
27. What does the abbreviation GM mean?
28. Define BM in terms of the second moment
of area I of the water plane, and the dis¬
placed volume V of a vessel.
29. What is the primary purpose of a ship’s in¬
clining experiment?
Part Four
Part Four
288 Mechanical Engineering Principles
Practice Exercise 133 Multiple-choice
questions on
hydrostatics
(Answers on page 336)
1. A force of 50 N acts uniformly over and
at right angles to a surface. When the area
of the surface is 5 m 2 , the pressure on the
area is:
(a) 250 Pa (b) 10 Pa
(c) 45 Pa (d) 55 Pa
2. Which of the following statements is false?
The pressure at a given depth in a fluid
(a) is equal in all directions
(b) is independent of the shape of the con¬
tainer
(c) acts at right angles to the surface con¬
taining the fluid
(d) depends on the area of the surface
3. A container holds water of density
1000 kg/m 3 . Taking the gravitational
acceleration as 10 m/s 2 , the pressure at a
depth of 100 mm is:
(a) 1 kPa (b) 1 MPa
(c) 100 Pa (d) 1 Pa
4. If the water in question 3 is now replaced
by a fluid having a density of 2000 kg/m 3 ,
the pressure at a depth of 100 mm is:
(a) 2 kPa (b) 500 kPa
(c) 200 Pa (d) 0.5 Pa
5. The gauge pressure of fluid in a pipe is
70 kPa and the atmospheric pressure is
100 kPa. The absolute pressure of the fluid
in the pipe is:
(a) 7 MPa (b) 30 kPa
(c) 170 kPa (d) 10/7 kPa
6. A (7-tube manometer contains mercury
of density 13600 kg/m 3 . When the differ¬
ence in the height of the mercury levels is
100 mm and taking the gravitational accel¬
eration as 10 m/s 2 , the gauge pressure is:
(a) 13.6 Pa (b) 13.6 MPa
(c) 13 710 Pa (d) 13.6 kPa
7. The mercury in the £/-tube of question 6
is to be replaced by water of density
1000 kg/m 3 . The height of the tube to con¬
tain the water for the same gauge pressure
is:
(a) (1/13.6) of the original height
(b) 13.6 times the original height
(c) 13.6 m more than the original height
(d) 13.6 m less than the original height
8. Which of the following devices does not
measure pressure?
(a) barometer (b) McLeod gauge
(c) thermocouple (d) manometer
9. A pressure of 10 kPa is equivalent to:
(a) 10 millibars (b) 1 bar
(c) 0.1 bar (d) 0.1 millibars
10. A pressure of 1000 mbars is equivalent to:
(a) 0.1 kN/m 2 (b) 10 kPa
(c) 1000 Pa (d) 100 kN/m 2
11. Which of the following statements is false?
(a) Barometers may be used for the mea¬
surement of altitude.
(b) Standard atmospheric pressure is the
pressure due to the mass of the air
above the ground.
(c) The maximum pressure that a mercury
manometer, using a 1 m length of glass
tubing, is capable of measuring is in
the order of 130 kPa.
(d) An inclined manometer is designed to
measure higher values of pressure than
the (7-tube manometer.
In questions 12 and 13 assume that atmospheric
pressure is 1 bar.
12. A Bourdon pressure gauge indicates a pres¬
sure of 3 bars. The absolute pressure of the
system being measured is:
(a) 1 bar (b) 2 bars
(c) 3 bars (d) 4 bars
13. In question 12, the gauge pressure is:
(a) 1 bar (b) 2 bars
(c) 3 bars (d) 4 bars
In questions 14 to 18 select the most suitable
pressure-indicating device from the following list:
(a) Mercury filled (7-tube manometer
(b) Bourdon gauge
(c) McLeod gauge
(d) aneroid barometer
Hydrostatics 289
(e) Pirani gauge
(f) Fortin barometer
(g) water-filled inclined barometer
14. A robust device to measure high pressures
in the range 0-30 MPa.
15. Calibration of a Pirani gauge.
16. Measurement of gas pressures comparable
with atmospheric pressure.
17. To measure pressures of the order of 1 MPa.
18. Measurement of atmospheric pressure to a
high degree of accuracy.
19. Figure 23.7(b), on page 280, shows a
CZ-tube manometer connected to a gas
under pressure. If atmospheric pressure is
76 cm of mercury and h { is measured in
centimetres then the gauge pressure (in cm
of mercury) of the gas is:
(a) h { (b) h { + 76
(c) /zj-76 (d) 76 -h x
20. In question 19 the absolute pressure of the
gas (in cm of mercury) is:
(a ) h { (b) h { + 76
(c) h { - 76 (d) 76 —Aj
21. Which of the following statements
is true?
(a) Atmospheric pressure of 101.325 kN/m 2
is equivalent to 101.325 millibars.
(b) An aneroid barometer is used as a stan¬
dard for calibration purposes.
(c) In engineering, ‘pressure’ is the force
per unit area exerted by fluids.
(d) Water is normally used in a barometer
to measure atmospheric pressure.
22. Which of the following statements is true
for a ship floating in equilibrium?
(a) The weight is larger than the buoyancy.
(b) The weight is smaller than the buoy¬
ancy.
(c) The weight is equal to the buoyancy.
(d) The weight is independent of the buoy¬
ancy.
23. For a ship to be initially stable, the meta-
centric height must be:
(a) positive
(b) negative
(c) zero
(d) equal to the buoyancy
24. For a ship to be stable, it is helpful if KG is:
(a) negative (b) large
(c) small (d) equal to KM
References
[1] ROSS, C.T.F. Pressure Vessels; External Pressure
Technology - 2nd Edition, Elsevier, Oxford, UK.
[2] Pressure - www.routledge.com/cw/bird
A video reference on YouTube on various types of
pressure, including atmospheric pressure, hydrostatic
pressure and wind pressure, which in the case of the
last type, was experienced by Hurricane Katrina & the
tornadoes experienced in Alabama and elsewhere.
[3] Hydrostatic Stability - www.routledge.com/cw/bird
A video reference on YouTube on the damage stability of
ro-ro ‘car’ ferries.
For fully worked solutions to each of the problems in Practice Exercises 126 to 133 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Four
Chapter 24
Fluid flow
Why it is important to understand: Fluid flow
The measurement of fluid flow is of great importance in many industrial processes, some examples
including air flow in the ventilating ducts of a coal mine, the flow rate of water in a condenser at a power
station, the flow rate of liquids in chemical processes, the control and monitoring of the fuel, lubricating
and cooling fluids of ships and aircraft engines, and so on. Fluid flow is one of the most difficult of
industrial measurements to carry out, since flow behaviour depends on a great many variables concerning
the physical properties of a fluid. There are available a large number of fluid flow measuring instruments
generally called flowmeters, which can measure the flow rate of liquids (in m 3 /s) or the mass flow rate of
gaseous fluids (in kg/s). The two main categories of flowmeters are differential pressure flowmeters and
mechanical flowmeters. This chapter also contains calculations on Bernoulli’s equation and the impact of
a jet on a stationary plate.
At the end of this chapter you should be able to:
• appreciate the importance of measurement of fluid flow
• describe the construction, principle of operation, advantages and disadvantages, and practical applications
of orifice plates, Venturi tubes, flow nozzles and Pitot-static tube and describe the principle of operation of
deflecting vane and turbine type flowmeters
• describe the construction, principle of operation, advantages and disadvantages, and practical applications of
float and tapered tube flowmeters, electromagnetic flowmeters, and hot-wire anemometer
• select the most appropriate flowmeter for a particular application
• state the continuity equation (i.e. the principle of conservation of mass)
• state and perform calculations on Bernoulli’s equation
• state and perform calculations on the impact of a jet on a stationary plate
24.1 Differential pressure flowmeters
When certain flowmeters are installed in pipelines they
often cause an obstruction to the fluid flowing in the
pipe by reducing the cross-sectional area of the pipeline.
This causes a change in the velocity of the fluid, with a
related change in pressure. Figure 24.1 shows a section
through a pipeline into which a flowmeter has been
inserted. The flow rate of the fluid may be determined
from a measurement of the difference between the
pressures on the walls of the pipe at specified distances
upstream and downstream of the flowmeter. Such
devices are known as differential pressure flowmeters.
The pressure difference in Figure 24.1 is measured
using a manometer connected to appropriate pressure
tapping points. The pressure is seen to be greater up¬
stream of the flowmeter than downstream, the pressure
difference being shown as h.
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
Fluid flow 291
Flowmeter causing obstruction in fluid flow
Calibration of the manometer depends on the shape of
the obstruction, the positions of the pressure tapping
points and the physical properties of the fluid.
In industrial applications the pressure difference is
detected by a differential pressure cell, the output from
which is either an amplified pressure signal or an elec¬
trical signal.
Examples of differential pressure flowmeters com¬
monly used include:
(a) Orifice plate (see Section 24.2)
(b) Venturi tube (see Section 24.3)
(c) Flow nozzles (see Section 24.4)
(d) Pitot-static tube (see Section 24.5)
British Standard reference BS 1042: Part 1: 1964 and
Part 2A: 1973 ‘Methods for the measurement of fluid
flow in pipes’ gives specifications for measurement,
manufacture, tolerances, accuracy, sizes, choice, and
so on, of differential flowmeters.
24.2 Orifice plate
Construction
An orifice plate consists of a circular, thin, flat plate with
a hole (or orifice) machined through its centre to fine
limits of accuracy. The orifice has a diameter less than
the pipeline into which the plate is installed and a typi¬
cal section of an installation is shown in Figure 24.2(a).
Orifice plates are manufactured in stainless steel, monel
metal, polyester glass fibre, and for large pipes, such as
sewers or hot gas mains, in brick and concrete.
Principles of operation
When a fluid moves through a restriction in a pipe, the
fluid accelerates and a reduction in pressure occurs, the
magnitude of which is related to the flow rate of the
fluid. The variation of pressure near an orifice plate is
shown in Figure 24.2(b). The position of minimum pres¬
sure is located downstream from the orifice plate where
Orifice plate
Figure 24.2
the flow stream is narrowest. This point of minimum
cross-sectional area of the jet is called the ‘vena con-
tracta’. Beyond this point the pressure rises but does not
return to the original upstream value and there is a per¬
manent pressure loss. This loss depends on the size and
type of orifice plate, the positions of the upstream and
downstream pressure tappings and the change in fluid
velocity between the pressure tappings that depends on
the flow rate and the dimensions of the orifice plate.
In Figure 24.2(a) corner pressure tappings are shown
at A and B. Alternatively, with an orifice plate insert¬
ed into a pipeline of diameter d , pressure tappings
are often located at distances of d and dll from the
plate respectively upstream and downstream. At
distance d upstream the flow pattern is not influenced
by the presence of the orifice plate, and distance dll
coincides with the vena contracta.
Advantages of orifice plates
(i) They are relatively inexpensive.
(ii) They are usually thin enough to fit between an
existing pair of pipe flanges.
Disadvantages of orifice plates
(i) The sharpness of the edge of the orifice can be¬
come worn with use, causing calibration errors.
(ii) The possible build-up of matter against the plate.
(iii) A considerable loss in the pumping efficiency
due to the pressure loss downstream of the plate.
Applications
Orifice plates are usually used in medium and large
pipes and are best suited to the indication and control
of essentially constant flow rates. Several applications
are found in the general process industries.
Part Four
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292 Mechanical Engineering Principles
24.3 Venturi tube
Construction
The Venturi tube or venturimeter is an instrument
for measuring with accuracy the flow rate of fluids
in pipes. A typical arrangement of a section through
such a device is shown in Figure 24.3, and consists
of a short converging conical tube called the inlet or
upstream cone, leading to a cylindrical portion called
the throat. A diverging section called the outlet or recov¬
ery cone follows this. The entrance and exit diameter is
the same as that of the pipeline into which it is installed.
Angle ft is usually a maximum of 21 °, giving a taper of
p/2 of 10.5°. The length of the throat is made equal to
the diameter of the throat. Angle a is about 5° to 7° to
ensure a minimum loss of energy but where this is
unimportant a can be as large as 14° to 15°.
Upstream
Figure 24.3
Pressure tappings are made at the entry (at A) and at
the throat (at B) and the pressure difference h which
is measured using a manometer, a differential pressure
cell or similar gauge, is dependent on the flow rate
through the meter. Usually pressure chambers are fitted
around the entrance pipe and the throat circumference
with a series of tapping holes made in the chamber to
which the manometer is connected. This ensures that
an average pressure is recorded. The loss of energy due
to turbulence that occurs just downstream with an ori¬
fice plate is largely avoided in the venturimeter due to
the gradual divergence beyond the throat.
Venturimeters are usually made a permanent instal¬
lation in a pipeline and are manufactured usually from
stainless steel, cast iron, monel metal or polyester glass
fibre.
Advantages of venturimeters
(i) High accuracy results are possible.
(ii) There is a low pressure loss in the tube (typically
only 2% to 3% in a well proportioned tube).
(iii) Venturimeters are unlikely to trap any matter
from the fluid being metered.
Disadvantages of venturimeters
(i) High manufacturing costs.
(ii) The installation tends to be rather long (typically
120 mm for a pipe of internal diameter 50 mm).
24.4 Flow nozzle
The flow nozzle lies between an orifice plate and the
venturimeter both in performance and cost. A typi¬
cal section through a flow nozzle is shown in Figure
24.4, where pressure tappings are located immediately
adjacent to the upstream and downstream faces of the
nozzle (i.e. at points A and B). The fluid flow does
not contract any further as it leaves the nozzle and
the pressure loss created is considerably less than that
occurring with orifice plates. Flow nozzles are suitable
for use with high velocity flows for they do not suffer
the wear that occurs in orifice plate edges during such
flows.
Figure 24.4
24.5 Pitot-static tube
A Pitot-static tube is a device for measuring the ve¬
locity of moving fluids or of the velocity of bodies
moving through fluids. It consists of one tube, called
the Pitot* tube, with an open end facing the direction
of the fluid motion, shown as pipe R in Figure 24.5,
and a second tube, called the piezometer tube, with
the opening at 90° to the fluid flow, shown as T in
Figure 24.5. Pressure recorded by a pressure gauge
moving with the flow, i.e. static or stationary rela¬
tive to the fluid, is called free stream pressure and
Fluid flow 293
Figure 24.5
connecting a pressure gauge to a small hole in the
wall of a pipe, such as point T in Figure 24.5, is the
easiest method of recording this pressure. The differ¬
ence in pressure (p R -p T ), shown as h in the manome¬
ter of Figure 24.5, is an indication of the speed of the
fluid in the pipe.
Figure 24.6 shows a practical Pitot-static tube con¬
sisting of a pair of concentric tubes. The centre tube
is the impact probe that has an open end which faces
‘head-on’ into the flow. The outer tube has a series of
holes around its circumference located at right angles
to the flow, as shown by A and B in Figure 24.6. The
manometer, showing a pressure difference of /z, may be
calibrated to indicate the velocity of flow directly.
Figure 24.6
Applications
A Pitot-static tube may be used for both turbulent
and noil-turbulent flow. The tubes can be made very
small compared with the size of the pipeline and the
monitoring of flow velocity at particular points in the
* ienri Pitot (3 May 1695-27 December 1771) was a French
hydraulic engineer and the inventor of the pitot tube. To find out
more go to www.routledge.com/cw/bird
cross-section of a duct can be achieved. The device is
generally unsuitable for routine measurements and in
industry is often used for making preliminary tests of
flow rate in order to specify permanent flow measuring
equipment for a pipeline. The main use of Pitot tubes is
to measure the velocity of solid bodies moving through
fluids, such as the velocity of ships. In these cases, the
tube is connected to a Bourdon pressure gauge that can
be calibrated to read velocity directly. A development
of the Pitot tube, a pitometer, tests the flow of water in
water mains and detects leakages.
Advantages of Pitot-static tubes
(i) They are inexpensive devices.
(ii) They are easy to install.
(iii) They produce only a small pressure loss in the
tube.
(iv) They do not interrupt the flow.
Disadvantages of Pitot-static tubes
(i) Due to the small pressure difference, they are
only suitable for high velocity fluids.
(ii) They can measure the flow rate only at a particu¬
lar position in the cross-section of the pipe.
(iii) They easily become blocked when used with
fluids carrying particles.
24.6 Mechanical flowmeters
With mechanical flowmeters, a sensing element situat¬
ed in a pipeline is displaced by the fluid flowing past it.
Examples of mechanical flowmeters commonly used
include:
(a) Deflecting vane flowmeter (see Section 24.7)
(b) Turbine type meters (see Section 24.8)
24.7 Deflecting vane flowmeter
The deflecting vane flowmeter consists basically of
a pivoted vane suspended in the fluid flow stream as
shown in Figure 24.7.
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294 Mechanical Engineering Principles
When a jet of fluid impinges on the vane it deflects
from its normal position by an amount proportional
to the flow rate. The movement of the vane is indi¬
cated on a scale that may be calibrated in flow units.
This type of meter is normally used for measuring
liquid flow rates in open channels or for measuring
the velocity of air in ventilation ducts. The main dis¬
advantages of this device are that it restricts the
flow rate and it needs to be recalibrated for fluids of
differing densities.
24.8 Turbine type meters
Turbine type flowmeters are those that use some
form of multi-vane rotor and are driven by the fluid
being investigated. Three such devices are the cup
anemometer, the rotary vane positive displacement
meter and the turbine flowmeter.
(a) Cup anemometer. An anemometer is an
instrument that measures the velocity of moving
gases and is most often used for the measurement
of wind speed. The cup anemometer has three or
four cups of hemispherical shape mounted at the
end of arms radiating horizontally from a fixed
point. The cup system spins round the vertical
axis with a speed approximately proportional
to the velocity of the wind. With the aid of a
mechanical and/or electrical counter the wind
speed can be determined and the device is easily
adapted for automatic recording.
(b) Rotary vane positive displacement meters
measure the flow rate by indicating the quantity
of liquid flowing through the meter in a given
time. A typical such device is shown in section in
Figure 24.8 and consists of a cylindrical chamber
into which is placed a rotor containing a number
of vanes (six in this case). Liquid entering the
chamber turns the rotor and a known amount of
liquid is trapped and carried round to the outlet.
If x is the volume displaced by one blade then
for each revolution of the rotor in Figure 24.8 the
total volume displaced is 6x. The rotor shaft may
be coupled to a mechanical counter and electri¬
cal devices which may be calibrated to give flow
volume. This type of meter in its various forms
is used widely for the measurement of domestic
and industrial water consumption, for the accu¬
rate measurement of petrol in petrol pumps and
for the consumption and batch control measure¬
ments in the general process and food industries
for measuring flows as varied as solvents, tar and
molasses (i.e. thickish treacle).
Figure 24.8
(c) A turbine flowmeter contains in its construc¬
tion a rotor to which blades are attached which
spin at a velocity proportional to the veloc¬
ity of the fluid which flows through the meter. A
typical section through such a meter is shown in
Figure 24.9. The number of revolutions made by
the turbine blades may be determined by a
mechanical or electrical device enabling the flow
rate or total flow to be determined. Advantages
of turbine flowmeters include a compact durable
form, high accuracy, wide temperature and pres¬
sure capability and good response characteristics.
Applications include the volumetric measurement
of both crude and refined petroleum products in
pipelines up to 600 mm bore, and in the water,
power, aerospace, process and food industries,
and with modification may be used for natural,
industrial and liquid gas measurements. Turbine
flowmeters require periodic inspection and clean¬
ing of the working parts.
Rotor
^
77777777777777777777777,
1777777777777777777777777777777A
Direction of
. fluid flow"
Figure 24.9
24.9 Float and tapered-tube meter
Principle of operation
With orifice plates and venturimeters the area of the
opening in the obstruction is fixed and any change
in the flow rate produces a corresponding change in
pressure. With the float and tapered-tube meter the
area of the restriction may be varied so as to maintain
a steady pressure differential. A typical meter of this
type is shown diagrammatically in Figure 24.10 where
Fluid flow 295
Figure 24.10
a vertical tapered tube contains a ‘float’ that has a den¬
sity greater than the fluid.
The float in the tapered tube produces a restriction
to the fluid flow. The fluid can only pass in the annular
area between the float and the walls of the tube. This
reduction in area produces an increase in velocity and
hence a pressure difference, which causes the float to
rise. The greater the flow rate, the greater is the rise in
the float position, and vice versa. The position of the
float is a measure of the flow rate of the fluid and this
is shown on a vertical scale engraved on a transpar¬
ent tube of plastic or glass. For air, a small sphere is
used for the float but for liquids there is a tendency
to instability and the float is then designed with vanes
that cause it to spin and thus stabilise itself as the
liquid flows past. Such meters are often called
‘rotameters’. Calibration of float and tapered tube
flowmeters can be achieved using a Pitot-static tube or,
more often, by using a weighing meter in an instrument
repair workshop.
Advantages of float and tapered-tube flowmeters
(i) They have a very simple design.
(ii) They can be made direct reading.
(iii) They can measure very low flow rates.
Disadvantages of float and tapered-tube flowmeters
(i) They are prone to errors, such as those caused by
temperature fluctuations.
(ii) They can only be installed vertically in a pipeline.
(iii) They cannot be used with liquids containing
large amounts of solids in suspension.
(iv) They need to be recalibrated for fluids of differ¬
ent densities.
Practical applications of float and tapered-tube
meters are found in the medical field, in instrument
purging, in mechanical engineering test rigs and in
simple process applications, in particular for very low
flow rates. Many corrosive fluids can be handled with
this device without complications.
24.10 Electromagnetic flowmeter
The flow rate of fluids that conduct electricity, such
as water or molten metal, can be measured using
an electromagnetic flowmeter whose principle of
operation is based on the laws of electromagnetic
induction. When a conductor of length L moves at
right angles to a magnetic field of density B at a velocity
v, an induced e.m.f. e is generated, given by: e = BLv.
With the electromagnetic flowmeter arrangement
shown in Figure 24.11, the fluid is the conductor and
the e.m.f. is detected by two electrodes placed across
the diameter of the non-magnetic tube.
Rearranging e = BLv gives:
e
velocity, v =
Figure 24.11
Thus with B and L known, when e is measured, the
velocity of the fluid can be calculated.
Main advantages of electromagnetic flowmeters
(i) Unlike other methods, there is nothing directly
to impede the fluid flow.
Part Four
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296 Mechanical Engineering Principles
(ii) There is a linear relationship between the fluid
flow and the induced e.m.f.
(iii) Flow can be metered in either direction by using
a centre-zero measuring instrument.
Applications of electromagnetic flowmeters are found
in the measurement of speeds of slurries, pastes and
viscous liquids, and they are also widely used in the
water production, supply and treatment industry.
24.11 Hot-wire anemometer
A simple hot-wire anemometer consists of a small piece
of wire which is heated by an electric current and po¬
sitioned in the air or gas stream whose velocity is to be
measured. The stream passing the wire cools it, the rate
of cooling being dependent on the flow velocity. In prac¬
tice there are various ways in which this is achieved:
(i) If a constant current is passed through the wire,
variation in flow results in a change of tempera¬
ture of the wire and hence a change in resistance
which may be measured by a Wheatstone bridge
arrangement. The change in resistance may be
related to fluid flow.
(ii) If the wire’s resistance, and hence temperature,
is kept constant, a change in fluid flow results in
a corresponding change in current which can be
calibrated as an indication of the flow rate.
(iii) A thermocouple may be incorporated in the as¬
sembly, monitoring the hot wire and recording
the temperature which is an indication of the air
or gas velocity.
Advantages of the hot-wire anemometer
(a) Its size is small.
(b) It has great sensitivity.
24.12 Choice of flowmeter
Choose the most appropriate
fluid flow measuring device for the following
circumstances:
(a) The most accurate, permanent installation for
measuring liquid flow rate.
(b) To determine the velocity of low-speed
aircraft and ships.
(c) Accurate continuous volumetric measurement
of crude petroleum products in a duct of
500 mm bore.
(d) To give a reasonable indication of the mean
flow velocity, while maintaining a steady
pressure difference on a hydraulic test rig.
(e) For an essentially constant flow rate with
reasonable accuracy in a large bore pipe, with
a cheap and simple installation.
(a) Venturimeter
(b) Pitot-static tube
(c) Turbine flowmeter
(d) Float and tapered-tube flowmeter
(e) Orifice plate.
Now try the following Practice Exercise
Practice Exercise 134 Further problems on
the measurement of
fluid flow
For the flow measurement devices listed 1 to 5,
(a) describe briefly their construction (b) state
their principle of operation (c) state their char¬
acteristics and limitations (d) state typical practi¬
cal applications (e) discuss their advantages and
disadvantages.
1. Orifice plate
2. Venturimeter
3. Pitot-static tube
4. Float and tapered-tube meter
5. Turbine flowmeter.
24.13 Equation of continuity
The calibrations of many of the flowmeters described
earlier are based on the equation of continuity and
Bernoulli’s equation.
The equation of continuity states that for the steady
flow of a fluid through a pipe of varying cross-section
the rate of mass entering the pipe must be equal to the
rate of mass leaving the pipe; this is really a statement
of the principle of conservation of mass. Thus, for an
incompressible fluid:
a l v l = a 2 v 2
Fluid flow 297
where a { = cross-sectional area at section 1,
a 2 = cross-sectional area at section 2, v { = velocity of
fluid at section 1, and v 2 = velocity of fluid at section 2
24.14 Bernoulli's equation
Bernoulli’s equation* states that for a fluid flowing
through a pipe from section 1 to section 2:
v;
7-+ L ^- + gZi = I ^-+ 1 T- + g(z 2 + hf)
v;
where p = density of the fluid,
P { = pressure at section 1,
P 2 = pressure at section 2,
* laniel Bernon (8 February 1700-17 March 1782) was a
Swiss mathematician and physicist who is most remembered
for his work in fluid mechanics and his pioneering work in
probability and statistics.
To find out more go to www.routledge.com/cw/bird
and
Vj = velocity at section 1,
v 2 = velocity at section 2,
Zj = ‘height’ of pipe at section 1,
z 2 = ‘height’ of pipe at section 2,
hf = friction losses (in m) due to the fluid flowing
from section 1 to section 2,
g = 9.81 m/s 2 (assumed).
A storage tank contains oil whose
free surface is 5 m above an outlet pipe, as shown
in Figure 24.12. Determine the mass rate of flow at
the exit of the outlet pipe, assuming that (a) losses
at the pipe entry = 0.4 v 2 , and (b) losses at the valve
= 0.25 v 2 .
Pipe diameter = 0.04 m, density of oil, p = 770 kg/m 3 .
Free surface
1 1
z 1 = 5 m
Figure 24.12
Let v 2 = velocity of oil through the outlet pipe.
From Bernoulli’s equation:
p
p 2
= — +
P
+ g* i
+ gz 2 + 0.4 v 2 + 0.25 v 2
i.e. 0 + 0 + g(5 m) = 0+ -y- + 0 + 0.65 v 2 2
(where in the above, the following assumptions have
been made: P { = P 2 = atmospheric pressure, and Vj is
negligible)
Hence, 5 m x 9.81 ^ = (0.5 + 0.65)v 2
s 2
2
9 m
Rearranging gives: 1.15 v 2 - 49.05 —y
s 2
Hence,
2 49.05
V 2 =
1.15
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298 Mechanical Engineering Principles
from which, v 2 =
Cross-sectional area of pipe = a 2 =
49.05
1.15
= 6.531 m/s
n x 0.04 2
4 4
= 0.001257 m 2
Mass rate of flow through the outlet pipe
= p a,v, = 770 x 1.257 x 10- 3 m 2 x 6.531
nr
- 6.321 kg/s
m
s
Flow through an orifice
Consider the flow of a liquid through a small orifice,
as shown in Figures 24.13(a) and (b), where it can be
seen that the vena contracta (VC) lies just to the right
of the orifice. The cross-sectional area of the fluid is the
smallest here and its decrease in area from the orifice is
measured by the coefficient of contraction (C).
Figure 24.13
Due to friction losses there will be a loss in velocity
at the orifice; this is measured by the coefficient of
velocity, namely C v , so that:
C d = C v x C c = the coefficient of discharge.
Let a = area of orifice.
Due to the vena contracta the equivalent cross-
sectional area = C c a
Now the theoretical velocity at section 2 = v 2 =^2gh
but due to friction losses, v 2 = C v sjlgh
and due to contraction, v 2 = C c C v ^2gh
Hence discharge = C Q a x C y ^2gh
But
Therefore,
C d =C v C c
discharge = C d x ayjlgh
Now try the following Practice Exercise
Practice Exercise 135 Further problems on
fluid flow
1. If in the storage tank of worked Problem 2 on
page 297, Figure 24.12, z { = 8 m, determine
the mass rate of flow from the outlet pipe.
[7.995 kg/s]
2. If in the storage tank of worked Problem 2,
page 297, Figure 24.12,z 1 = 10 m, determine
the mass rate of flow from the outlet pipe.
[8.939 kg/s]
3. If in Figure 24.13, h = 6 m, C, = 0.8,
C y = 0.7, determine the values of C d and
actual v 2 . [C d = 0.56, v 2 = 6.08 m/s]
4. If in Figure 24.13, h = 10 m, C = 0.75,
C v = 0.65, and the cross-sectional area is
1.5 x 10 -3 m 2 , determine the discharge and
the actual velocity v 2 .
\C d = 0.488, 6.84 m/s, 7.90 kg/s]
24.15 Impact of a jet on a
stationary plate
The impact of a jet on a plate is of importance in a
number of engineering problems, including the deter¬
mination of pressures on buildings subjected to gusts
of wind.
Consider the jet of fluid acting on the flat plate of
Figure 24.14, where it can be seen that the velocity of
the fluid is turned through 90°, or change of velocity = v.
Figure 24.14
Fluid flow 299
Now, momentum = mv and as v is constant, the change
of momentum =
dm
dm
dt
x v
However, — = mass rate of flow = pav
dt
Therefore, change of momentum = pav x v = pav 2 but
from Newton’s second law of motion (see pages 183
and 189),
F = rate of change of momentum
i.e. F = pav 2
where F = resulting normal force on the flat plate.
force oav 2 ^
Pressure =-= ±_- = pv 2
area a
For wide surfaces, such as garden fences, the pressure
can be calculated by the above formula, but for tall
buildings and trees, civil engineers normally assume
that:
Pressure p = 0.5 pv 2
This is because the flow of fluid is similar to the plan
view shown in Figure 24.15, where the change of mo¬
mentum is much less.
Figure 24.15
Determine the wind pressure on a
slim, tall building due to a gale of 100 km/h. Take
density of air, p= 1.2 kg/m 3 .
For a tall building, pressure p — 0.5 pv 2
km 1000 m lh
Velocity, v = 100 x —-x - - = 27.78 m/s
: h km 3600s
Hence,
kg
/
wind pressure, p = 0.5 x 1.2 — T x
m
27.78 —
v
/
= 462.96 N/m 2 = 0.00463 bar.
What would be the wind pressure of
Problem 3, if the gale were acting on a very wide
and flat surface?
For a very wide surface,
pressure,/; = pv 2 = 1.2
kg
m
x
\2
27.78 —
v /
= 926.1 N/m 2 = 0.00926 bar
(or less than 1/100th of atmospheric pressure!)
Now try the following Practice Exercises
Practice Exercise 136 Further problems on
the impact of jets on
flat surfaces
1. A hurricane of velocity 220 km/h blows
perpendicularly on to a very wide flat surface.
Determine the wind pressure that acts on
this surface due to this hurricane, when the
density of air, p = 1.2 kg/m 3 . [0.0448 bar]
2. What is the wind pressure for Problem 1 on
a slim, tall building? [0.0224 bar]
3. A tornado with a velocity of 320 km/h blows
perpendicularly on to a very wide surface.
Determine the wind pressure that acts on this
surface due to this tornado, when the density
of air, p = 1.23 kg/m 3 . [0.0972 bar]
4. What is the wind pressure for Problem 3 on
a slim, tall building? [0.0486 bar]
5. If atmospheric pressure were 1.014 bar, what
fraction of atmospheric pressure would be
the wind pressure calculated in Problem 4?
0.0479
Practice Exercise 137 Short-answer
questions on the
measurement of
fluid flow
In the flowmeters listed 1 to 10, state typical
practical applications of each.
1. Orifice plate.
2. Venturimeter.
3. Float and tapered-tube meter.
4. Electromagnetic flowmeter.
5. Pitot-static tube.
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300 Mechanical Engineering Principles
6. Hot-wire anemometer.
7. Turbine flowmeter.
8. Deflecting vane flowmeter.
9. Flow nozzles.
10. Rotary vane positive displacement meter.
11. Write down the relationship between the
coefficients C c , C v and C d
12. Write down the formula for the pressure
due to a wind acting perpendicularly on a
tall slender building.
Practice Exercise 138 Multiple-choice
questions on the
measurement of
fluid flow
(Answers on page 336)
1. The term ‘flow rate’ usually refers to:
(a) mass flow rate
(b) velocity of flow
(c) volumetric flow rate
2. The most suitable method for measuring the
velocity of high-speed gas flow in a duct is:
(a) venturimeter
(b) orifice plate
(c) Pitot-static tube
(d) float and tapered-tube meter
3. Which of the following statements is false?
When a fluid moves through a restriction in
a pipe, the fluid
(a) accelerates and the pressure increases
(b) decelerates and the pressure decreases
(c) decelerates and the pressure increases
(d) accelerates and the pressure decreases
4. With an orifice plate in a pipeline the vena
contracta is situated:
(a) downstream at the position of mini¬
mum cross-sectional area of flow
(b) upstream at the position of minimum
cross-sectional area of flow
(c) downstream at the position of maxi¬
mum cross-sectional area of flow
(d) upstream at the position of maximum
cross-sectional area of flow
In questions 5 to 14, select the most appropriate
device for the particular requirements from the
following list:
(a) orifice plate
(b) turbine flowmeter
(c) flow nozzle
(d) pitometer
(e) venturimeter
(f) cup anemometer
(g) electromagnetic flowmeter
(h) pitot-static tube
(i) float and tapered-tube meter
(j) hot-wire anemometer
(k) deflecting vane flowmeter.
5. Easy to install, reasonably inexpensive, for
high-velocity flows.
6. To measure the flow rate of gas, incorporat¬
ing a Wheatstone bridge circuit.
7. Very low flow rate of corrosive liquid in a
chemical process.
8. To detect leakages from water mains.
9. To determine the flow rate of liquid metals
without impeding its flow.
10. To measure the velocity of wind.
11. Constant flow rate, large bore pipe, in the
general process industry.
12. To make a preliminary test of flow rate in
order to specify permanent flow measuring
equipment.
13. To determine the flow rate of fluid very ac¬
curately with low pressure loss.
14. To measure the flow rate of air in a ventilat¬
ing duct.
15. For a certain wind velocity, what fraction
of the pressure would act on a tall slender
building in comparison with a very wide
surface?
(a) 0.01 (b) 0
(c) 0.5 (d) 0.99
16. For a wind speed of 190 km/h, what frac¬
tion (approximate) of atmospheric pressure
will this be, when blowing perpendicularly
to a very wide surface?
(a) 2.5 (b) 0.5
(c) 1/30 (d) 0
For fully worked solutions to each of the problems in Practice Exercises 134 to 138 in this chapter,
go to the website:
www.routledge.com/cw/bird
Chapter 25
Why it is important to understand: Ideal gas laws
The relationships that exist between pressure, volume and temperature in a gas are given in a set of laws
called the gas laws, the most fundamental being those of Boyle’s, Charles’, and the pressure law, together
with Dalton’s law of partial pressures and the characteristic gas equation. These laws are used for all
sorts of practical applications, including for designing pressure vessels, in the form of circular cylinders
and spheres, which are used for storing and transporting gases. Another example of this is the pressure
in car tyres, which can increase due to a temperature increase, and can decrease due to a temperature
decrease. Other examples are large and medium size gas storage cylinders and domestic spray cans,
which can explode if they are heated. In the case of domestic spray cans, these can explode dangerously
in a domestic situation if they are left on a window sill where the sunshine acting on them causes them
to heat up, or if they are thrown on to a fire. In these cases, the consequence can be disastrous, so don’t
throw your ‘full’ spray can on to a fire; you may very sadly and deeply regret it! Another example of a gas
storage vessel is that used by your ‘local’ gas companies, which supply natural gas (methane) to domestic
properties, businesses, etc.
At the end of this chapter you should be able to:
• state and perform calculations involving Boyle’s law
• understand the term isothermal
• state and perform calculations involving Charles’ law
• understand the term isobaric
• state and perform calculations involving the pressure law
• state and perform calculations on Dalton’s law of partial pressures
• state and perform calculations on the characteristic gas equation
• understand the term STP
25.1 Boyle's law
1
Boyle’s law* states:
i.e.
P oc -
F V
the volume V of a fixed mass of gas is inversely
proportional to its absolute pressure p at con¬
or
k
P ~ V
stant temperature
or
pV= k at constant temperature, where
Mechanical Engineering Principles, Bird and Ross, ISBN 9780415517850
Part Four
302 Mechanical Engineering Principles
p = absolute pressure in pascals (Pa),
V= volume in m 3 , and k = a constant.
Changes that occur at constant temperature are called
isothermal changes. When a fixed mass of gas at con¬
stant temperature changes from pressure p x and vol¬
ume V l to pressure p 2 and volume V x then:
P\ V \ = Pl V 2
A gas occupies a volume of 0.10 m 3
at a pressure of 1.8 MPa. Determine (a) the pressure
if the volume is changed to 0.06 m 3 at constant
temperature, and (b) the volume if the pressure is
changed to 2.4 MPa at constant temperature.
(a) Since the change occurs at constant temperature
(i.e. an isothermal change), Boyle’s law applies,
i.e. P\ v \=Pi V 2
where p x = 1.8 MPa, V i = 0.10 m 3 and
V 2 = 0.06 m 3 .
Hence, (1.8)(0.10) =p 2 (0.06)
r , • , 1.8x0.10 „ „„
from which, pressure p 2 = —— = 3 MPa
* Robert Boyle (25 January 1627-31 December 1691) is
regarded as the first modem chemist and is best known for
Boyle’s law.
To find out more go to www.routledge.com/cw/bird
(b) P\V y =P 2 V 2 where/-! = 1.8 MPa, Kj= 0.10 m 3
and p 2 = 2.4 MPa.
Hence, (1.8)(0.10) = (2.4) F 2
from which, volume K =-= 0.075 rrr
2 2.4
In an isothermal process, a mass of
gas has its volume reduced from 3200 mm 3 to
2000 mm 3 . If the initial pressure of the gas is
110 kPa, determine the final pressure.
Since the process is isothermal, it takes place at con¬
stant temperature and hence Boyle’s law applies, i.e.
p x V x =p 2 V 2 , where p 2 = 110 kPa, V { = 3200 mm 3 and
V 2 = 2000 mm 3 .
Hence, (110)(3200) = p 2 (2000)
110x 3200
from which, final pressure, p 2 = ———r—
\J
= 176 kPa
Some gas occupies a volume of
1.5 m 3 in a cylinder at a pressure of 250 kPa. A
piston, sliding in the cylinder, compresses the gas
isothermally until the volume is 0.5 m 3 . If the area
of the piston is 300 cm 2 , calculate the force on the
piston when the gas is compressed.
An isothermal process means constant temperature and
thus Boyle’s law applies, \.q. p x V x = p 2 V 2
where V x = 1.5 m 3 , V 2 = 0.5 m 3 and p x = 250 kPa.
Hence, (250)(1.5) =/? 2 (0.5)
250x 1.5
from which, pressure, p 1 = ———
z 0.5
- 750 kPa
force
Pressure =-, from which, force = pressure x area.
area
Hence, force on the piston
= (750 x 10 3 Pa)(300 x 10^ m 2 )
= 22.5 kN
The gas in a syringe has a pressure of
600 mm of mercury (Hg) and a volume of quantity
20 mL. If the syringe is compressed to a volume
of 3 mL, what will be the pressure of the gas,
assuming that its temperature does not change?
Since P\V\ = P 2^2
Ideal gas laws 303
then (600 mm Hg)(20 mL) = (p 2 )(3 mL)
from which, new pressure,
600 mm Hg x 20 mL
p 2 = = 4000 mm of mercury
Now try the following Practice Exercise
Practice Exercise 139 Further problems on
Boyle's law
1. The pressure of a mass of gas is increased
from 150 kPa to 750 kPa at constant tem¬
perature. Determine the final volume of the
gas, if its initial volume is 1.5 m 3 . [0.3 m 3 ]
2. In an isothermal process, a mass of gas has
its volume reduced from 50 cm 3 to 32 cm 3 .
If the initial pressure of the gas is 80 kPa,
determine its final pressure. [125 kPa]
3.
The piston of an air compressor compresses
air to
1
4
of its original volume during its
stroke. Determine the final pressure of the air
if the original pressure is 100 kPa, assuming
an isothermal change. [400 kPa]
4. A quantity of gas in a cylinder occupies a
volume of 2 m 3 at a pressure of 300 kPa. A
piston slides in the cylinder and compresses
the gas, according to Boyle’s law, until the
volume is 0.5 m 3 . If the area of the piston
is 0.02 m 2 , calculate the force on the piston
when the gas is compressed. [24 kN]
5. The gas in a simple pump has a pressure of
400 mm of mercury (Hg) and a volume of
10 mL. If the pump is compressed to a vol¬
ume of 2 mL, calculate the pressure of the
gas, assuming that its temperature does not
change.
[2000 mm of mercury]
25.2 Charles'law
Charles’ law* states:
for a given mass of gas at constant pressure, the
volume V is directly proportional to its thermo¬
dynamic temperature T
i.e. foe r
or V=kT
V ,
or — = k at constant pressure, where
T = thermodynamic temperature in
kelvin (K).
A process that takes place at constant pressure is called
an isobaric process.
The relationship between the Celsius scale of tem¬
perature and the thermodynamic or absolute scale is
given by:
kelvin = degrees Celsius + 273
i.e. K = °C + 273
or °C = K - 273 (as stated in Chapter 21).
If a given mass of gas at a constant pressure occupies
a volume V l at a temperature T l and a volume V 2 at
temperature T 2 , then
A gas occupies a volume of
1.2 litres at 20°C. Determine the volume it
occupies at 130°C if the pressure is kept constant.
Since the change occurs at constant pressure (i.e. an
isobaric process), Charles’ law applies,
Yi = h.
7
* Jacques Alexandre Cesar Charles (12 November 1767-7
April 1823) was a French inventor, scientist, mathematician
and balloonist. With his brother, Charles launched the world’s
first hydrogen filled balloon and is also well known for Charles’
law, which describes how gases expand when heated. To find
out more go to www.routledge.com/cw/bird
Part Four
Part Four
304 Mechanical Engineering Principles
where V 1 = 1.2 litre, 7\ = 20°C = (20 + 273)K = 293 K
Now try the following Practice Exercise
and T 2 = (130 + 273)K = 403 K.
Hence,
H = l2_
293 403
from which, volume at 130°C, V 2 =
(1-2X403)
293
= 1.65 litres
Gas at a temperature of 150°C has
its volume reduced by one-third in an isobaric
process. Calculate the final temperature of the gas.
Since the process is isobaric it takes place at constant
pressure and hence Charles’ law applies,
i.e.
where
and
Hence
T\ T 2
r, =( 150 + 273)K = 423 K
V,=
— V
3
2
l
C,
l _ 3
Vy
423 T,
from which, final temperature,
2
T 2 = -(423) (423) = 282 K
3
or (282 - 273)°C i.e. 9°C
A balloon is under a constant internal
pressure. If its volume is 15 litres at a temperature
of 35°C, what will be its volume at a temperature
of 10°C.
Since
V\ _*2
Tx T 2
where
T l = (35 + 273)K = 308 K
and
T 2 = (10 + 273)K = 283 K
Hence,
15 litres V 2
308K 283K
and
new volume, V 2 =
15 litres x283K _ 0 ,.,
- 13.8 litres
308K
Practice Exercise 140 Further problems on
Charles' law
1. Some gas initially at 16°C is heated to 96°C
at constant pressure. If the initial volume of
the gas is 0.8 m 3 , determine the final volume
of the gas. [1.02 m 3 ]
2. A gas is contained in a vessel of volume
0.02 m 3 at a pressure of 300 kPa and a tem¬
perature of 15°C. The gas is passed into a
vessel of volume 0.015 m 3 . Determine to
what temperature the gas must be cooled for
the pressure to remain the same. [-57°C]
3. In an isobaric process gas at a temperature
of 120°C has its volume reduced by a sixth.
Determine the final temperature of the gas.
[54.5°C]
4. The volume of a balloon is 30 litres at a tem¬
perature of 27°C. If the balloon is under a
constant internal pressure, calculate its vol¬
ume at a temperature of 12°C. [28.5 litres]
25.3 The pressure or Gay-Lussac's law
The pressure or Gay-Lussac’s law states:
the pressure p of a fixed mass of gas is directly
proportional to its thermodynamic temperature
T at constant volume .
i.e. p oc T or p = kT or ^ = k
When a fixed mass of gas at constant volume changes
from pressure p { and temperature T v to pressure p 2 and
temperature T 2 then:
Ei = Pi
Gas initially at a temperature of 17°C
and pressure 150 kPa is heated at constant volume
until its temperature is 124°C. Determine the final
pressure of the gas, assuming no loss of gas.
Since the gas is at constant volume, the pressure law
r . P\ _ Pi
applies, i.e. — - —
1 \ L i
Ideal gas laws 305
where
T { = (17 + 273)K = 290 K,
T 2 = (124 + 273)K = 397 K
and
p x = 150 kPa
Hence
from which,
290 397
, . (150)(397)
final pressure, p 2 = ——-
= 205.3 kPa
A rigid pressure vessel is subjected to
a gas pressure of 10 atmospheres at a temperature of
20°C. The pressure vessel can withstand a maximum
pressure of 30 atmospheres. What gas temperature
increase will this vessel withstand?
(Note that 1 atmosphere of pressure means
1.01325 bar or 1.01325 x 10 * 1 2 * * 5 Pa or 14.5 psi)
P_\_ _ Pi_
Ti T 2
T { = (20 + 273)K = 293 K,
10 atmospheres _ 30 atmospheres
293K ~ T 2
from which, new temperature,
Since
where
then
25.4 Dalton's law of partial pressure
Dalton’s law* of partial pressure states:
the total pressure of a mixture of gases
occupying a given volume is equal to the sum of
the pressures of each gas , considered separately,
at constant temperature .
The pressure of each constituent gas when occupying
a fixed volume alone is known as the partial pressure
of that gas.
An ideal gas is one that completely obeys the gas
laws given in Sections 25.1 to 25.4. In practice no gas
is an ideal gas, although air is very close to being one.
For calculation purposes the difference between an
ideal and an actual gas is very small.
A gas R in a container exerts a
pressure of 200 kPa at a temperature of 18°C.
Gas Q is added to the container and the pressure
increases to 320 kPa at the same temperature.
Determine the pressure that gas Q alone exerts at
the same temperature.
r, - — atmospheresX 293 K _ 879 Kor (g7 „ _ 273) „ c
10 atmospheres
= 606°C
Hence, temperature rise = (879 - 293)K = 586 K
or temperature rise = (606 - 20)°C = 586°C
Note that a temperature change of 586 K is equal to a
temperature change of 586°C
Now try the following Practice Exercise
Practice Exercise 141 Further problems
on the pressure law
1. Gas, initially at a temperature of 27°C and
pressure 100 kPa, is heated at constant
volume until its temperature is 150°C. As¬
suming no loss of gas, determine the final
pressure of the gas. [141 kPa]
2. A pressure vessel is subj ected to a gas pressure
of 8 atmospheres at a temperature of 15°C. The
vessel can withstand a maximum pressure of
28 atmospheres. Calculate the gas temperature
increase the vessel can withstand. [720°C]
* lohn Dalton (6 September 1766-27 July 1844) is best known
for his pioneering work in the development of modem atomic
theory and his research into colour blindness. To find out more
go to www.routledge.com/cw/bird
Part Four
Part Four
306 Mechanical Engineering Principles
Initial pressure,^ = 200 kPa, and the pressure of gases
R and Q together, p = p R + Pq = 320 kPa
By Dalton’s law of partial pressure, the pressure of
gas Q alone is
Pq = P ~ Pr = 320 - 200 = 120 kPa.
Now try the following Practice Exercise
Practice Exercise 142 A further problem
on Dalton's law of
partial pressure
1. A gas A in a container exerts a pressure of
120 kPa at a temperature of 20°C. Gas B is
added to the container and the pressure in¬
creases to 300 kPa at the same temperature.
Determine the pressure that gas B alone
exerts at the same temperature. [180 kPa]
25.5 Characteristic gas equation
Frequently, when a gas is undergoing some change, the
pressure, temperature and volume all vary simultane¬
ously. Provided there is no change in the mass of a gas,
the above gas laws can be combined, giving
P^l . — P^l . = k where k is a constant.
Tx T 2
For an ideal gas, constant k = mR , where m is the
mass of the gas in kg, and R is the characteristic gas
. pV
constant, i.e. = mR
or
T
pV= mRT
This is called the characteristic gas equation. In this
equation, p = absolute pressure in pascals,
V= volume in m 3 , m = mass in kg,
R = characteristic gas constant in J/(kg K),
and T= thermodynamic temperature in kelvin.
Some typical values of the characteristic gas constant
R include:
air, 287 J/(kg K), hydrogen 4160 J/(kg K), oxygen
260 J/(kg K) and carbon dioxide 184 J/(kg K).
Standard temperature and pressure (i.e. STP)
refers to a temperature of 0°C, i.e. 273 K, and normal
atmospheric pressure of 101.325 kPa.
25.6 Worked problems on the
characteristic gas equation
A gas occupies a volume of 2.0 m 3
when at a pressure of 100 kPa and a temperature of
120°C. Determine the volume of the gas at 15°C if
the pressure is increased to 250 kPa.
Using the combined gas law:
. P\ V \ _ P2 V 2
T,
T-
where V 1 = 2.0 m 3 ,p 1 = 100 kPa ,p 2 = 250 kPa,
T j = (120 + 273)K = 393 K and
T 2 = (15 + 273) K = 288 K, gives:
(100)(2.0) _ (250)F 2
393 “ 288
from which, volume at 15°C,
(100)(2.0)(288)
(393)(250)
= 0.586 m 3
20000 mm 3 of air initially at a
pressure of 600 kPa and temperature 180°C is
expanded to a volume of 70000 mm 3 at a pressure
of 120 kPa. Determine the final temperature of the
air, assuming no losses during the process.
Using the combined gas law:
Eh
r,
Eh
t 2
where V 1 = 20000 mm 3 ,
V 2 = 70000 mm 3 ,£>j = 600 kPa,
p 2 = 120 kPa, and
T l = (180 + 273) K = 453 K
Hence (600)(20000) = (120) (70000)
453 T 2
from which, final temperature,
(120)(70000)(453) =31?K Qr ^
2 (600)(20000)
Some air at a temperature of 40°C
and pressure 4 bar occupies a volume of 0.05 m 3 .
Determine the mass of the air assuming the
characteristic gas constant for air to be 287 J/(kg K).
From above, pV= mRT,
where p = 4 bar = 4 x 10 5 Pa
(since 1 bar = 10 5 Pa - see Chapter 23),
Ideal gas laws 307
V= 0.05 m 3 ,
T= (40 + 273) K= 313 K,
and R = 287 J/(kg K).
Hence (4 x 10 5 )(0.05) = w(287)(313)
, . (4 x 10 5 )(0.05)
irom which, mass ot air, m = - -—--
(287)(313)
= 0.223 kg or 223 g
A cylinder of helium has a volume
of 600 cm 3 . The cylinder contains 200 g of helium
at a temperature of 25 °C. Determine the pressure
of the helium if the characteristic gas constant for
helium is 2080 J/(kg K).
From the characteristic gas equation, pV= mRT\
V= 600 cm 3 = 600 x 10 -6 m 3 , m = 200 g = 0.2 kg,
T= (25 + 273) K = 298 K and R = 2080 J/(kg K)
Hence (/?)(600 x 10" 6 ) = (0.2)(2080)(298)
from which, pressure, p
(0.2)(2080)(298)
(600 x 10 -6 )
= 206613333 Pa
= 206.6 MPa
A spherical vessel has a diameter of
1.2 m and contains oxygen at a pressure of 2 bar
and a temperature of -20°C. Determine the mass
of oxygen in the vessel. Take the characteristic gas
constant for oxygen to be 0.260 kJ/(kg K).
From the characteristic gas equation, pV= mRT ',
V = volume of spherical vessel
and
Hence,
4 3 4
= — nr = — n
r \3
1.2
v 2 ,
5
= 0.905 m 3 ,
3 3
p = 2 bar = 2 x 10^ Pa,
T= (-20 + 273) K = 253 K
R = 0.260 kJ/(kg K) = 260 J/(kg K)
(2 x 10 5 )(0.905) = m(26 0)(253)
from which, mass of oxygen, m
= (2 x 10 5 )(0.905)
(260X253)
= 2.75 kg
Determine the characteristic gas con¬
stant of a gas which has a specific volume of 0.5 m 3 /
kg at a temperature of 20°C and pressure 150 kPa.
From the characteristic gas equation, pV= mRT
pV
from which,
R =
mT
where p = 150 x 10 3 Pa,
T= (20 + 273) K = 293 K and
specific volume, Vim = 0.5 m 3 /kg.
Hence the characteristic gas constant,
fl-
/
R =
\ ( ^
150 x10 3
v
/
293
(0.5)
v /
256 J/(kg K)
Now try the following Practice Exercise
Practice Exercise 143 Further problems
on the characteristic
gas equation
1. A gas occupies a volume of 1.20 m 3 when
at a pressure of 120 kPa and a temperature
of 90°C. Determine the volume of the gas at
20°C if the pressure is increased to 320 kPa.
[0.363 m 3 ]
2. A given mass of air occupies a volume of
0.5 m 3 at a pressure of 500 kPa and a tem¬
perature of 20°C. Find the volume of the air
at STP [2.30 m 3 ]
3. A balloon is under an internal pressure of
110 kPa with a volume of 16 litres at a tem¬
perature of 22°C. If the balloon’s internal
pressure decreases to 50 kPa, what will be its
volume if the temperature decreases to 12°C.
[34.0 litres]
4. A spherical vessel has a diameter of
2.0 m and contains hydrogen at a pressure of
300 kPa and a temperature of -30°C. De¬
termine the mass of hydrogen in the vessel.
Assume the characteristic gas constant R for
hydrogen is 4160 J/(kg K). [1.24 kg]
5. A cylinder 200 mm in diameter and
1.5 m long contains oxygen at a pressure of
2 MPa and a temperature of 20°C. Determine
the mass of oxygen in the cylinder. Assume
the characteristic gas constant for oxygen is
260 J/(kgK). [1.24 kg]
6 . A gas is pumped into an empty cylinder of
volume 0.1 m 3 until the pressure is 5 MPa.
The temperature of the gas is 40°C. If the
Part Four
Part Four
308 Mechanical Engineering Principles
cylinder mass increases by 5.32 kg when the
gas has been added, determine the value of
the characteristic gas constant.
[300 J/(kg K)]
7. The mass of a gas is 1.2 kg and it occupies
a volume of 13.45 m 3 at STP. Determine its
characteristic gas constant.
8 .
[4160 J/(kg K)]
30 cm 3 of air initially at a pressure of
500 kPa and temperature 150°C is expand¬
ed to a volume of 100 cm 3 at a pressure of
200 kPa. Determine the final temperature
of the air, assuming no losses during the
process. [291°C]
A quantity of gas in a cylinder occupies a
volume of 0.05 m 3 at a pressure of 400 kPa
and a temperature of 27°C. It is compressed
according to Boyle’s law until its pressure
is 1 MPa, and then expanded according to
Charles’ law until its volume is 0.03 m 3 .
Determine the final temperature of the gas.
[177°C]
10. Some air at a temperature of 35°C and pres¬
sure 2 bar occupies a volume of 0.08 m 3 .
Determine the mass of the air assuming
the characteristic gas constant for air to be
287 J/(kg K). (1 bar = 10 5 Pa) [0.181 kg]
11. Determine the characteristic gas constant
R of a gas that has a specific volume of
0.267 m 3 /kg at a temperature of 17°C and
pressure 200 kPa. [ 184 J/(kg K)]
25.7 Further worked problems on the
characteristic gas equation
A vessel has a volume of 0.80 m 3
and contains a mixture of helium and hydrogen at
a pressure of 450 kPa and a temperature of 17°C.
If the mass of helium present is 0.40 kg determine
(a) the partial pressure of each gas, and (b) the mass
of hydrogen present. Assume the characteristic gas
constant for helium to be 2080 J/(kg K) and for
hydrogen 4160 J/(kg K).
(a) V = 0.80 m 3 , p = 450 kPa, T = (17 + 273)K =
290 K, m He = 0.40 kg, R He = 2080 J/(kg K).
If p He is the partial pressure of the helium, then
using the characteristic gas equation,
PHe V=m He R He T g* VeS:
(P/feX0-80) = (0.40)(2080)(290)
from which, the partial pressure of the helium,
(0.40K2080X290) =30L6kpa
He (0.80)
By Dalton’s law of partial pressure the total pres¬
sure p is given by the sum of the partial pressures,
i.e. p = p H + p He , from which,
the partial pressure of the hydrogen,
p H = P ~P He = 450 - 301.6 = 148.4 kPa
(b) From the characteristic gas equation,
Ph V= m H R H T
Hence, (148.4 x 10 3 )(0.8) = m i/ (4160)(290)
from which, mass of hydrogen,
(148.4 x 10 3 )(0.8)
m
H
(4160)(290)
0.098 kg or 98 g
A compressed air cylinder has a
volume of 1.2 m 3 and contains air at a pressure of
1 MPa and a temperature of 25°C. Air is released
from the cylinder until the pressure falls to 300 kPa
and the temperature is 15°C. Determine (a) the mass
of air released from the container, and (b) the volume
it would occupy at STP. Assume the characteristic
gas constant for air to be 287 J/(kg K).
v i = 1.2 m 3 (= V 2 ),p 1 = 1 MPa = 10 6 Pa,
T, = (25 + 273)K = 298 K,
T 2 = (15 + 273)K = 288 K,
p 2 = 300 kPa = 300 x 10 3 Pa
and R = 287 J/(kg K)
(a) Using the characteristic gas equation,
P\V\ = m x RT x , to find the initial mass of air
in the cylinder gives:
(10 6 )(1.2) = m j(287)(298)
from which, massm, = = 14.03 kg
1 (287)(298)
Similarly, using p ? V 2 = m 2^2 t0 ^ ie
mass of air in the cylinder gives:
(300 x 10 3 )( 1.2) = ot 2 (287)(288)
Ideal gas laws 309
from which, mass m 9 =
(300 x 10 3 )(1.2)
(287X288)
4.36 kg
Mass of air released from cylinder
= m | — m 2 — 14.03 - 4.36 = 9.67 kg.
(b) At STP, T- 273 K and p = 101.325 kPa.
Using the characteristic gas equation pV= mRT
mRT
volume, V= -
P
(9.67)(287)(273)
101325
= 7.48 m 3
A vessel X contains gas at a
pressure of 750 kPa at a temperature of 27°C. It
is connected via a valve to vessel Y that is filled
with a similar gas at a pressure of 1.2 MPa and a
temperature of 27°C. The volume of vessel X is
2.0 m 3 and that of vessel Y is 3.0 m 3 . Determine
the final pressure at 27°C when the valve is opened
and the gases are allowed to mix. Assume R for the
gas to be 300 J/(kg K).
For vessel A:
p x = 750 x 10 3 Pa, T x = (27 + 273)K = 300 K,
V x = 2.0 m 3 and R = 300 J/(kg K)
From the characteristic gas equation, p x V x = m x RT x
Hence (750 x 10 3 )(2.0) = m x (300)(300)
from which, mass of gas in vessel X,
(750 x 10 3 )(2.0)
(3001(300) " 16 67 kg -
For vessel Y:
p Y = 1.2 x 10 6 Pa, T y = (27 + 273)K = 300 K,
V } ,= 3.0 m 3 and R = 300 J/(kg K)
From the characteristic gas equation, p Y V Y = m } RT y
Hence (1.2 x 10 6 )(3.0) = m y (300)(300)
from which, mass of gas in vessel Y,
_ (1.2x 10 6 )(3.0)
mY (300)(300) 40 kg
When the valve is opened, mass of mixture,
m = m x + m Y = 16.67 + 40 = 56.67 kg.
Total volume, V= V x + V y =2.0 + 3.0 = 5.0 m 3 ,
R -300 J/(kg K), T= 300 K.
From the characteristic gas equation, pV= mRT
p( 5.0) = (56.67)(300)(300)
from which, final pressure,
(56.67)(300)(300)
5.0
= 1.02 MPa
Now try the following Practice Exercises
Practice Exercise 144 Further questions
on ideal gas laws
1. A vessel P contains gas at a pressure of
800 kPa at a temperature of 25°C. It is con¬
nected via a valve to vessel Q that is filled
with similar gas at a pressure of 1.5 MPa
and a temperature of 25°C. The volume of
vessel P is 1.5 m 3 and that of vessel R is
2.5 m 3 . Determine the final pressure at 25°C
when the valve is opened and the gases are
allowed to mix. Assume R for the gas to be
297 J/(kgK). [1.24 MPa]
2. A vessel contains 4 kg of air at a pressure
of 600 kPa and a temperature of 40°C. The
vessel is connected to another by a short pipe
and the air exhausts into it. The final pres¬
sure in both vessels is 250 kPa and the tem¬
perature in both is 15°C. If the pressure in
the second vessel before the air entered was
zero, determine the volume of each vessel.
Assume R for air is 287 J/(kg K).
[0.60 m 3 , 0.72 m 3 ]
3. A vessel has a volume of 0.75 m 3 and con¬
tains a mixture of air and carbon dioxide at
a pressure of 200 kPa and a temperature of
27°C. If the mass of air present is 0.5 kg
determine (a) the partial pressure of each
gas and (b) the mass of carbon dioxide. As¬
sume the characteristic gas constant for air
to be 287 J/(kg K) and for carbon dioxide
184 J/(kg K).
[(a) 57.4 kPa, 142.6 kPa (b) 1.94 kg]
4. A mass of gas occupies a volume of 0.02 m 3
when its pressure is 150 kPa and its tempera¬
ture is 17°C. If the gas is compressed until
its pressure is 500 kPa and its temperature
is 57°C, determine (a) the volume it will oc¬
cupy and (b) its mass, if the characteristic
gas constant for the gas is 205 J/(kg K).
[(a) 0.0068 m 3 (b) 0.050 kg]
Part Four
Part Four
310 Mechanical Engineering Principles
5. A compressed air cylinder has a volume
of 0.6 m 3 and contains air at a pressure of
1.2 MPa absolute and a temperature of 37°C.
After use the pressure is 800 kPa absolute
and the temperature is 17°C. Calculate
(a) the mass of air removed from the cylinder,
and (b) the volume the mass of air removed
would occupy at STP conditions. Take R for
air as 287 J/(kg K) and atmospheric pressure
as 100 kPa.
[(a) 2.33 kg (b) 1.83 m 3 ]
Practice Exercise 145 Short-answer
questions on ideal
gas laws
1. State Boyle’s law.
2. State Charles’ law.
3. State the Pressure law.
4. State Dalton’s law of partial pressures.
5. State the relationship between the Celsius
and the thermodynamic scale of temperature.
6. What is (a) an isothermal change, and (b) an
isobaric change?
7. Define an ideal gas.
8. State the characteristic gas equation.
9. What is meant by STP?
Practice Exercise 146 Multiple-choice
questions on ideal
gas laws
(Answers on page 336)
1. Which of the following statements is
false?
(a) At constant temperature, Charles’ law
applies.
(b) The pressure of a given mass of gas
decreases as the volume is increased at
constant temperature.
(c) Isobaric changes are those which occur
at constant pressure.
(d) Boyle’s law applies at constant tem¬
perature.
2. A gas occupies a volume of 4 m 3 at a pres¬
sure of 400 kPa. At constant temperature,
the pressure is increased to 500 kPa. The
new volume occupied by the gas is:
(a) 5 m 3 (b) 0.3 m 3
(c) 0.2 m 3 (d) 3.2 m 3
3. A gas at a temperature of 27°C occupies a
volume of 5 m 3 . The volume of the same
mass of gas at the same pressure but at a
temperature of 57°C is:
(a) 10.56 m 3 (b) 5.50 m 3
(c) 4.55 m 3 (d) 2.37 m 3
4. Which of the following statements is
false?
(a) An ideal gas is one that completely
obeys the gas laws.
(b) Isothermal changes are those that occur
at constant volume.
(c) The volume of a gas increases when
the temperature increases at constant
pressure.
(d) Changes that occur at constant pressure
are called isobaric changes.
A gas has a volume of 0.4 m 3 when its pressure
is 250 kPa and its temperature is 400 K. Use this
data in questions 5 and 6.
5. The temperature when the pressure is
increased to 400 kPa and the volume is in¬
creased to 0.8 m 3 is:
(a) 400 K (b) 80 K
(c) 1280 K (d) 320 K
6. The pressure when the temperature is
raised to 600 K and the volume is reduced
to 0.2 m 3 is:
(a) 187.5 kPa (b) 250 kPa
(c) 333.3 kPa (d) 750 kPa
7. A gas has a volume of 3 m 3 at a temperature
of 546 K and a pressure of 101.325 kPa.
The volume it occupies at STP is:
(a) 3 m 3 (b) 1.5 m 3
(c) 6 m 3
8. Which of the following statements is false?
(a) A characteristic gas constant has units
of J/(kg K).
(b) STP conditions are 273 K and
101.325 kPa.
Ideal gas laws 311
(c) All gases are ideal gases.
(d) An ideal gas is one that obeys the gas
laws.
A mass of 5 kg of air is pumped into a con¬
tainer of volume 2.87 m 3 . The characteristic gas
constant for air is 287 J/(kg K). Use this data in
questions 9 and 10.
9. The pres sure when the temperature is 2 7 °C is:
(a) 1.6 kPa (b) 6 kPa
(c) 150 kPa (d) 15kPa
10. The temperature when the pressure is
200 kPa is:
(a) 400°C (b) 127°C
(c) 127 K (d) 283 K
For fully worked solutions to each of the problems in Practice Exercises 139 to 146 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Four
Chapter 26
Why it is important to understand: The measurement of temperature
A change in temperature of a substance can often result in a change in one or more of its physical
properties. Thus, although temperature cannot be measured directly, its effects can be measured. Some
properties of substances used to determine changes in temperature include changes in dimensions,
electrical resistance, state, type and volume of radiation and colour.
Temperature measuring devices available are many and varied. Those described in this chapter are
those most often used in science and industry. The measurement of temperature is important in medi¬
cines and very many branches of science and engineering.
At the end of this chapter you should be able to:
• describe the construction, principle of operation and practical applications of the following temperature
measuring devices:
(a) liquid-in-glass thermometer (including advantages of mercury, and sources of error)
(b) thermocouples (including advantages and sources of error)
(c) resistance thermometer (including limitations and advantages of platinum coil)
(d) thermistors
(e) pyrometers (total radiation and optical types, including advantages and disadvantages)
• describe the principle of operation of
(a) temperature indicating paints and crayons
(b) bimetallic thermometers
(c) mercury-in-steel thermometer
(d) gas thermometer
• select the appropriate temperature measuring device for a particular application.
26.1 Liquid-in-glass thermometer
A liquid-in-glass thermometer uses the expansion of
a liquid with increase in temperature as its principle of
operation.
Construction
A typical liquid-in-glass thermometer is shown in Fig¬
ure 26.1 and consists of a sealed stem of uniform small¬
bore tubing, called a capillary tube, made of glass, with
a cylindrical glass bulb formed at one end. The bulb
and part of the stem are filled with a liquid such as mer-
Mechanical Engineering Principles. Bird and Ross, ISBN 9780415517850
The measurement of temperature 313
cury or alcohol and the remaining part of the tube is
evacuated. A temperature scale is formed by etching
graduations on the stem. A safety reservoir is usually
provided, into which the liquid can expand without
bursting the glass if the temperature is raised beyond
the upper limit of the scale.
Safety
Liquid Capillary tube reservoir
Bulb 0 10 20 30 40 50 60 70 80 90 100
Scale
Figure 26.1
Principle of operation
The operation of a liquid-in-glass thermometer depends
on the liquid expanding with increase in temperature
and contracting with decrease in temperature. The
position of the end of the column of liquid in the tube
is a measure of the temperature of the liquid in the
bulb - shown as 15°C in Figure 26.1, which is about
room temperature. Two fixed points are needed to
calibrate the thermometer, with the interval between
these points being divided into ‘degrees’. In the first
thermometer, made by Celsius, the fixed points chosen
were the temperature of melting ice (0°C) and that of
boiling water at standard atmospheric pressure (100°C),
in each case the blank stem being marked at the liquid
level. The distance between these two points, called the
fundamental interval, was divided into 100 equal parts,
each equivalent to 1°C, thus forming the scale.
The clinical thermometer, with a limited scale
around body temperature, the maximum and/or
minimum thermometer, recording the maximum
day temperature and minimum night temperature, and
the Beckman thermometer*, which is used only in
accurate measurement of temperature change, and has
no fixed points, are particular types of liquid-in-glass
thermometer which all operate on the same principle.
Advantages
The liquid-in-glass thermometer is simple in construc¬
tion, relatively inexpensive, easy to use and portable,
and is the most widely used method of temperature
measurement having industrial, chemical, clinical and
meteorological applications.
Disadvantages
Liquid-in-glass thermometers tend to be fragile and
hence easily broken, can only be used where the liquid
column is visible, cannot be used for surface tempera¬
ture measurements, cannot be read from a distance and
are unsuitable for high temperature measurements.
Advantages of mercury
The use of mercury in a thermometer has many advan¬
tages, for mercury:
(i) is clearly visible,
(ii) has a fairly uniform rate of expansion,
(iii) is readily obtainable in the pure state,
(iv) does not ‘wet’ the glass,
(v) is a good conductor of heat.
Mercury has a freezing point of -39°C and cannot
be used in a thermometer below this temperature. Its
boiling point is 357°C but before this temperature is
reached some distillation of the mercury occurs if the
space above the mercury is a vacuum. To prevent this,
and to extend the upper temperature limits to over
500°C, an inert gas such as nitrogen under pressure is
used to fill the remainder of the capillary tube. Alcohol,
often dyed red to be seen in the capillary tube, is con¬
siderably cheaper than mercury and has a freezing point
of-113°C, which is considerably lower than for mer¬
cury. However it has a low boiling point at about 79°C.
*Eri st Otto Beckmann (4 July 1843-12 July 1923) was a
German chemist best known for his invention of the Beckmann
differential thermometer and for his discoveiy of the Beckmann
rearrangement.
To find out more go to www.routledge.com/cw/bird
Part Four
Part Four
314 Mechanical Engineering Principles
Errors
Typical errors in liquid-in-glass thermometers may oc¬
cur due to:
(i) the slow cooling rate of glass
(ii) incorrect positioning of the thermometer
(iii) a delay in the thermometer becoming steady (i.e.
slow response time)
(iv) non-uniformity of the bore of the capillary tube,
which means that equal intervals marked on the
stem do not correspond to equal temperature
intervals.
26.2 Thermocouples
Thermocouples use the e.m.f. set up when the junction
of two dissimilar metals is heated.
Principle of operation
At the junction between two different metals, say, cop¬
per and constantan, there exists a difference in electri¬
cal potential, which varies with the temperature of the
junction. This is known as the Thermo-electric effect’. If
the circuit is completed with a second junction at a dif¬
ferent temperature, a current will flow round the circuit.
This principle is used in the thermocouple. Two different
metal conductors having their ends twisted together are
shown in Figure 26.2. If the two junctions are at different
temperatures, a current I flows round the circuit.
Metal A Metal B
Y
Figure 26.2
The deflection on the galvanometer G depends on the
difference in temperature between junctions X and Y
and is caused by the difference between voltages V x
and V. The higher temperature junction is usually
called the ‘hot junction’ and the lower temperature
junction the ‘cold junction’. If the cold junction is kept
at a constant known temperature, the galvanometer
can be calibrated to indicate the temperature of the hot
junction directly. The cold junction is then known as
the reference junction.
In many instrumentation situations, the measur¬
ing instrument needs to be located far from the point
at which the measurements are to be made. Extension
leads are then used, usually made of the same material
as the thermocouple but of smaller gauge. The refer¬
ence junction is then effectively moved to their ends.
The thermocouple is used by positioning the hot junc¬
tion where the temperature is required. The meter will
indicate the temperature of the hot junction only if the
reference junction is at 0°C for:
(temperature of hot junction) = (temperature of the
cold junction) + (temperature difference)
In a laboratory the reference junction is often placed
in melting ice, but in industry it is often positioned in
a thermostatically controlled oven or buried under¬
ground where the temperature is constant.
Construction
Thermocouple junctions are made by twisting together
the ends of two wires of dissimilar metals before
welding them. The construction of a typical copper-
constantan thermocouple for industrial use is shown
in Figure 26.3. Apart from the actual junction the
two conductors used must be insulated electrically
from each other with appropriate insulation and is
shown in Figure 26.3 as twin-holed tubing. The wires
and insulation are usually inserted into a sheath for
protection from environments in which they might be
damaged or corroded.
Hot junction
Figure 26.3
The measurement of temperature 315
Applications
A copper-constantan thermocouple can measure tem¬
perature from -250°C up to about 400°C, and is used
typically with boiler flue gases, food processing and
with sub-zero temperature measurement. An iron-
constantan thermocouple can measure temperature
from -200°C to about 850°C, and is used typically in
paper and pulp mills, re-heat and annealing furnaces
and in chemical reactors. A chromel-alumel thermo¬
couple can measure temperatures from -200°C to
about 1100°C and is used typically with blast furnace
gases, brick kilns and in glass manufacture.
For the measurement of temperatures above 1100°C
radiation pyrometers are normally used. However,
thermocouples are available made of platinum-
platinum/rhodium, capable of measuring temperatures
up to 1400°C, or tungsten-molybdenum which can
measure up to 2600°C.
Advantages
A thermocouple:
(i) has a very simple, relatively inexpensive con¬
struction
(ii) can be made very small and compact
(iii) is robust
(iv) is easily replaced if damaged
(v) has a small response time
(vi) can be used at a distance from the actual mea¬
suring instrument and is thus ideal for use with
automatic and remote-control systems.
Sources of error
Sources of error in the thermocouple, which are dif¬
ficult to overcome, include:
(i) voltage drops in leads and junctions
(ii) possible variations in the temperature of the cold
junction
(iii) stray thermoelectric effects, which are caused
by the addition of further metals into the ‘ideal’
two-metal thermocouple circuit.
Additional leads are frequently necessary for extension
leads or voltmeter terminal connections.
A thermocouple may be used with a battery- or
mains-operated electronic thermometer instead of a
millivoltmeter. These devices amplify the small e.m.f.s
from the thermocouple before feeding them to a multi¬
range voltmeter calibrated directly with temperature
scales. These devices have great accuracy and are
almost unaffected by voltage drops in the leads and
junctions.
A chromel-alumel thermocouple
generates an e.m.f. of 5 mV. Determine the tem¬
perature of the hot junction if the cold junction is
at a temperature of 15°C and the sensitivity of the
thermocouple is 0.04 mV/°C.
5 mV
Temperature difference for 5 mV = 7 - 7 -— T
F 0.04 mV/ C
= 125°C.
Temperature at hot junction = temperature of cold
junction + temperature difference
- 15°C + 125°C = 140°C
Now try the following Practice Exercise
Practice Exercise 147 Further problem on
the thermocouple
1. A platinum-platinum/rhodium thermocouple
generates an e.m.f. of 7.5 mV. If the cold
junction is at a temperature of 20°C, deter¬
mine the temperature of the hot junction.
Assume the sensitivity of the thermocouple
to be 6 pV/°C. [1270°C]
26.3 Resistance thermometers
Resistance thermometers use the change in electrical
resistance caused by temperature change.
Construction
Resistance thermometers are made in a variety of sizes,
shapes and forms depending on the application for which
they are designed. A typical resistance thermometer
is shown diagrammatically in Figure 26.4. The most
common metal used for the coil in such thermometers is
platinum even though its sensitivity is not as high as other
metals such as copper and nickel. However, platinum is
a very stable metal and provides reproducible results
in a resistance thermometer. A platinum resistance
thermometer is often used as a calibrating device. Since
platinum is expensive, connecting leads of another
metal, usually copper, are used with the thermometer to
connect it to a measuring circuit.
The platinum and the connecting leads are shown
joined at A and B in Figure 26.4, although some¬
times this junction may be made outside of the sheath.
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316 Mechanical Engineering Principles
Copper
dummy leads
Copper
extension
leads
(such as mica or
ceramic tubing)
Figure 26.4
However, these leads often come into close contact
with the heat source which can introduce errors into
the measurements. These may be eliminated by includ¬
ing a pair of identical leads, called dummy leads, which
experience the same temperature change as the exten¬
sion leads.
Principle of operation
With most metals a rise in temperature causes an
increase in electrical resistance, and since resistance
can be measured accurately this property can be used
to measure temperature. If the resistance of a length of
wire at 0°C is R 0 , and its resistance at 0 °C is R (j , then
Rq = Rq (1 + aO ), where a is the temperature coefficient
of resistance of the material.
Rq — Rq
Rearranging gives: temperature, 6 = ~—
aK o
Values of Rq and a may be determined experimental¬
ly or obtained from existing data. Thus, if R 0 can be
measured, temperature 0 can be calculated. This is the
principle of operation of a resistance thermometer. Al¬
though a sensitive ohmmeter can be used to measure
R 0 , for more accurate determinations a Wheatstone
bridge circuit is used as shown in Figure 26.5. This
circuit compares an unknown resistance R 0 with others
of known values, R x and R 2 being fixed values and R 3
being variable. Galvanometer G is a sensitive centre-
zero microammeter. R 3 is varied until zero deflection
is obtained on the galvanometer, i.e. no current flows
through G and the bridge is said to be ‘balanced’.
At balance: R 2 Rq = R 1 R 3
from which,
^ 1^3
*2
and if R { and R ? are of equal value, then R 0 = R 3
A resistance thermometer may be connected between
points A and B in Figure 26.5 and its resistance R 0 at
any temperature 6 accurately measured. Dummy leads
included in arm BC help to eliminate errors caused by
the extension leads which are normally necessary in
such a thermometer.
Limitations
Resistance thermometers using a nickel coil are used
mainly in the range -100°C to 300°C, whereas plati¬
num resistance thermometers are capable of measuring
with greater accuracy temperatures in the range -200°C
to about 800°C. This upper range may be extended to
about 1500°C if high melting point materials are used
for the sheath and coil construction.
Advantages and disadvantages of a platinum coil
Platinum is commonly used in resistance thermom¬
eters since it is chemically inert, i.e. un-reactive, re¬
sists corrosion and oxidation and has a high melting
point of 1769°C. A disadvantage of platinum is its slow
response to temperature variation.
The measurement of temperature 317
Applications
Platinum resistance thermometers may be used as
calibrating devices or in applications such as heat-
treating and annealing processes and can be adapted
easily for use with automatic recording or control
systems. Resistance thermometers tend to be fragile and
easily damaged especially when subjected to excessive
vibration or shock.
A platinum resistance thermometer
has a resistance of 25 Q at 0°C. When measuring
the temperature of an annealing process a
resistance value of 60 Q is recorded. To what
temperature does this correspond? Take the
temperature coefficient of resistance of platinum as
0.0038/°C
Figure 26.6. The resistance of a typical thermistor can
vary from 400 Q at 0°C to 100 Q at 140°C.
R 0 = R 0 ( 1 + aO), where R 0 = 25 Q,
R 0 = 60 Q and a = 0.0038/°C.
Rq — Rq
Rearranging gives: temperature, 0 = --—
aK o
60-25
(0.003 8)(25)
= 368.4°C
Figure 26.6
Advantages
The main advantages of a thermistor are its high sen¬
sitivity and small size. It provides an inexpensive
method of measuring and detecting small changes in
temperature.
Now try the following Practice Exercise
Exercise 148 Further problem on the
resistance thermometer
1. A platinum resistance thermometer has a
resistance of 100 Q at 0°C. When measur¬
ing the temperature of a heat process a re¬
sistance value of 177 Q is measured using a
Wheatstone bridge. Given that the tempera¬
ture coefficient of resistance of platinum is
0.0038/°C, determine the temperature of the
heat process, correct to the nearest degree.
[203°C]
26.4 Thermistors
A thermistor is a semi-conducting material - such
as mixtures of oxides of copper, manganese, cobalt,
etc. - in the form of a fused bead connected to two
leads. As its temperature is increased its resistance rap¬
idly decreases. Typical resistance/temperature curves
for a thermistor and common metals are shown in
26.5 Pyrometers
A pyrometer is a device for measuring very high tem¬
peratures and uses the principle that all substances emit
radiant energy when hot, the rate of emission depend¬
ing on their temperature. The measurement of thermal
radiation is therefore a convenient method of determin¬
ing the temperature of hot sources and is particularly
useful in industrial processes. There are two main types
of pyrometer, namely the total radiation pyrometer and
the optical pyrometer.
Pyrometers are very convenient instruments since
they can be used at a safe and comfortable distance
from the hot source. Thus applications of pyrometers
are found in measuring the temperature of molten
metals, the interiors of furnaces or the interiors of
volcanoes. Total radiation pyrometers can also be
used in conjunction with devices which record and
control temperature continuously.
Total radiation pyrometer
A typical arrangement of a total radiation pyrometer is
shown in Figure 26.7. Radiant energy from a hot source,
such as a furnace, is focused on to the hot junction of
a thermocouple after reflection from a concave mirror.
The temperature rise recorded by the thermocouple
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318 Mechanical Engineering Principles
Mirror
Hot
source
Figure 26.7
Filament lamp
Telescope
arrangement
Red filter
Hot
source
Radiation from
hot source
(such as a furnace) Variable
resistor
Figure 26.8
depends on the amount of radiant energy received, which
in turn depends on the temperature of the hot source.
The galvanometer G shown connected to the thermo¬
couple records the current which results from the e.m.f.
developed and may be calibrated to give a direct reading
of the temperature of the hot source. The thermocouple
is protected from direct radiation by a shield as shown
and the hot source may be viewed through the sighting
telescope. For greater sensitivity, a thermopile may be
used, a thermopile being a number of thermocouples
connected in series. Total radiation pyrometers are used
to measure temperature in the range 700°C to 2000°C.
Optical pyrometers
When the temperature of an object is raised sufficient¬
ly, two visual effects occur; the object appears brighter
and there is a change in colour of the light emitted.
These effects are used in the optical pyrometer where
a comparison or matching is made between the bright¬
ness of the glowing hot source and the light from a fila¬
ment of known temperature.
The most frequently used optical pyrometer is the dis¬
appearing filament pyrometer and a typical arrangement
is shown in Figure 26.8. A filament lamp is built into a
telescope arrangement which receives radiation from a
hot source, an image of which is seen through an eye¬
piece. A red filter is incorporated as a protection to the eye.
The current flowing through the lamp is controlled
by a variable resistor. As the current is increased, the
temperature of the filament increases and its colour
changes. When viewed through the eyepiece the
filament of the lamp appears superimposed on the image
of the radiant energy from the hot source. The current
is varied until the filament glows as brightly as the
background. It will then merge into the background and
seem to disappear. The current required to achieve this
is a measure of the temperature of the hot source and
the ammeter can be calibrated to read the temperature
directly. Optical pyrometers may be used to measure
temperatures up to, and even in excess of, 3000°C.
Advantages of pyrometers
(i) There is no practical limit to the temperature that
a pyrometer can measure.
(ii) A pyrometer need not be brought directly into
the hot zone and so is free from the effects of
The measurement of temperature 319
heat and chemical attack that can often cause
other measuring devices to deteriorate in use.
(iii) Very fast rates of change of temperature can be
followed by a pyrometer.
(iv) The temperature of moving bodies can be mea¬
sured.
(v) The lens system makes the pyrometer virtually
independent of its distance from the source.
Disadvantages of pyrometers
(i) A pyrometer is often more expensive than other
temperature measuring devices.
(ii) A direct view of the heat process is necessary.
(iii) Manual adjustment is necessary.
(iv) A reasonable amount of skill and care is required
in calibrating and using a pyrometer. For each
new measuring situation the pyrometer must be
re-calibrated.
(v) The temperature of the surroundings may affect
the reading of the pyrometer and such errors are
difficult to eliminate.
26.6 Temperature indicating
paints and crayons
Temperature indicating paints contain substances
which change their colour when heated to certain
temperatures. This change is usually due to chemi¬
cal decomposition, such as loss of water, in which the
change in colour of the paint after having reached the
particular temperature will be a permanent one. How¬
ever, in some types the original colour returns after
cooling. Temperature indicating paints are used where
the temperature of inaccessible parts of apparatus and
machines is required. They are particularly useful in
he at-treatment processes where the temperature of
the component needs to be known before a quenching
operation. There are several such paints available and
most have only a small temperature range so that dif¬
ferent paints have to be used for different temperatures.
The usual range of temperatures covered by these
paints is from about 30°C to 700°C.
Temperature sensitive crayons consist of fusible
solids compressed into the form of a stick. The melt¬
ing point of such crayons is used to determine when a
given temperature has been reached. The crayons are
simple to use but indicate a single temperature only,
i.e. its melting point temperature. There are over 100
different crayons available, each covering a particular
range of temperature. Crayons are available for tem¬
peratures within the range of 50°C to 1400°C. Such
crayons are used in metallurgical applications such as
preheating before welding, hardening, annealing or
tempering, or in monitoring the temperature of critical
parts of machines or for checking mould temperatures
in the rubber and plastics industries.
26.7 Bimetallic thermometers
Bimetallic thermometers depend on the expansion
of metal strips which operate an indicating pointer.
Two thin metal strips of differing thermal expansion
are welded or riveted together and the curvature of the
bimetallic strip changes with temperature change. For
greater sensitivity the strips may be coiled into a flat
spiral or helix, one end being fixed and the other being
made to rotate a pointer over a scale. Bimetallic ther¬
mometers are useful for alarm and over-temperature
applications where extreme accuracy is not essential.
If the whole is placed in a sheath, protection from cor¬
rosive environments is achieved but with a reduction
in response characteristics. The normal upper limit of
temperature measurement by this thermometer is about
200°C, although with special metals the range can be
extended to about 400°C.
26.8 Mercury-in-steel thermometer
The mercury-in-steel thermometer is an extension
of the principle of the mercury-in-glass thermome¬
ter. Mercury in a steel bulb expands via a small bore
capillary tube into a pressure indicating device, say a
Bourdon gauge, the position of the pointer indicating
the amount of expansion and thus the temperature. The
advantages of this instrument are that it is robust and,
by increasing the length of the capillary tube, the gauge
can be placed some distance from the bulb and can thus
be used to monitor temperatures in positions which are
inaccessible to the liquid-in-glass thermometer. Such
thermometers may be used to measure temperatures up
to 600°C.
26.9 Gas thermometers
The gas thermometer consists of a flexible U-tube of
mercury connected by a capillary tube to a vessel con¬
taining gas. The change in the volume of a fixed mass
of gas at constant pressure, or the change in pressure of
a fixed mass of gas at constant volume, may be used to
measure temperature. This thermometer is cumbersome
Part Four
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320 Mechanical Engineering Principles
and rarely used to measure temperature directly, but it
is often used as a standard with which to calibrate other
types of thermometer. With pure hydrogen the range
of the instrument extends from -240°C to 1500°C and
measurements can be made with extreme accuracy.
26.10 Choice of measuring devices
State which device would be most
suitable to measure the following:
(a) metal in a furnace, in the range 50°C to 1600°C
(b) the air in an office in the range 0°C to 40°C
(c) boiler flue gas in the range 15°C to 300°C
(d) a metal surface, where a visual indication is
required when it reaches 425°C
(e) materials in a high-temperature furnace in the
range 2000°C to 2800°C
(f) to calibrate a thermocouple in the range
-100°C to 500°C
(g) brick in a kiln up to 900°C
(h) an inexpensive method for food processing
applications in the range -25°C to -75°C.
(a) Radiation pyrometer
(b) Mercury-in-glass thermometer
(c) Copper-constantan thermocouple
(d) Temperature sensitive crayon
(e) Optical pyrometer
(f) Platinum resistance thermometer or gas ther¬
mometer
(g) Chromel-alumel thermocouple
(h) Alcohol-in-glass thermometer.
Now try the following Practice Exercises
Practice Exercise 149 Short-answer
questions on the
measurement of
temperature
For each of the temperature measuring devices
listed in 1 to 10, state very briefly its principle of
operation and the range of temperatures that it is
capable of measuring.
1. Mercury-in-glass thermometer.
2. Alcohol-in-glass thermometer.
3. Thermocouple.
4. Platinum resistance thermometer.
5. Total radiation pyrometer.
6. Optical pyrometer.
7. Temperature sensitive crayons.
8. Bimetallic thermometer.
9. Mercury-in-steel thermometer.
10. Gas thermometer.
Practice Exercise 150 Multiple-choice
questions on the
measurement of
temperature
(Answers on page 336)
1. The most suitable device for measuring
very small temperature changes is a
(a) thermopile (b) thermocouple
(c) thermistor
2. When two wires of different metals are
twisted together and heat applied to the
junction, an e.m.f. is produced. This effect
is used in a thermocouple to measure:
(a) e.m.f. (b) temperature
(c) expansion (d) heat
3. A cold junction of a thermocouple is at
room temperature of 15°C. A voltmeter
connected to the thermocouple circuit
indicates 10 mV. If the voltmeter is cali¬
brated as 20°C/mV, the temperature of
the hot source is:
(a) 185°C (b) 200°C
(c) 35°C (d) 215°C
4. The e.m.f. generated by a copper-
constantan thermometer is 15 mV. If the
cold junction is at a temperature of 20°C,
the temperature of the hot junction when
the sensitivity of the thermocouple is
0.03 mV/°C is:
The measurement of temperature 321
(a) 480°C (b) 520°C
(c) 20.45°C (d) 500°C
In questions 5 to 12, select the most appropriate
temperature measuring device from this list.
(a) copper-constantan thermocouple
(b) thermistor
(c) mercury-in-glass thermometer
(d) total radiation pyrometer
(e) platinum resistance thermometer
(f) gas thermometer
(g) temperature sensitive crayon
(h) alcohol-in-glass thermometer
(i) bimetallic thermometer
(j) mercury-in-steel thermometer
(k) optical pyrometer.
5. Over-temperature alarm at about 180°C.
6. Food processing plant in the range -250°C
to +250°C.
7. Automatic recording system for a
heat treating process in the range 90°C to
250°C.
8. Surface of molten metals in the range
1000°C to 1800°C.
9. To calibrate accurately a mercury-in-glass
thermometer.
10. Furnace up to 3000°C.
11. Inexpensive method of measuring very
small changes in temperature.
12. Metal surface where a visual indication
is required when the temperature reaches
520°C.
For fully worked solutions to each of the problems in Practice Exercises 147 to 150 in this chapter,
go to the website:
www.routledge.com/cw/bird
Part Four
Part Four
Revision Test 9 Hydrostatics, fluid flow, gas laws and temperature measurement
This Revision Test covers the material contained in Chapters 23 to 26. The marks for each question are shown in
brackets at the end of each question.
When required take the density of water to be 1000 kg/m 3 and gravitational acceleration as 9.81 m/s 2 .
1. A circular piston exerts a pressure of 150 kPa
on a fluid when the force applied to the piston is
0.5 kN. Calculate the diameter of the piston,
correct to the nearest millimetre. (6)
2. A tank contains water to a depth of 500 mm.
Determine the water pressure
(a) at a depth of 300 mm, and
(b) at the base of the tank. (6)
3. When the atmospheric pressure is 101 kPa,
calculate the absolute pressure, to the nearest
kilopascal, at a point on a submarine which is
50 m below the sea water surface. Assume that
the density of sea water is 1030 kg/m 3 . (5)
4. A body weighs 2.85 N in air and 2.35 N when
completely immersed in water. Determine
(a) the volume of the body
(b) the density of the body, and
(c) the relative density of the body. (9)
5. A submarine dives to a depth of 700 m. What is
the gauge pressure on its surface if the density of
seawater is 1020 kg/m 3 and g = 9.81 m/s 2 . (5)
6. State the most appropriate fluid flow measuring
device for the following applications:
(a) A high accuracy, permanent installation, in an
oil pipeline.
(b) For high velocity chemical flow, which does
not suffer wear.
(c) To detect leakage in water mains.
(d) To measure petrol in petrol pumps.
(e) To measure the speed of a viscous liquid.
(5)
7. A storage tank contains water to a depth of 7 m
above an outlet pipe, as shown in Figure 24.12
on page 297. The system is in equilibrium until
a valve in the outlet pipe is opened. Determine
the initial mass rate of flow at the exit of the out¬
let pipe, assuming that losses at the pipe entry
= 0.3 v 2 , and losses at the valve = 0.2 v 2 . The pipe
diameter is 0.05 m and the water density, p, is
1000 kg/m 3 . (15)
8. Determine the wind pressure acting on a slender
building due to a gale of 150 km/h that acts per¬
pendicularly to the building. Take the density of
air as 1.23 kg/m 3 . (5)
9. Some gas occupies a volume of 2.0 m 3 in a cylin¬
der at a pressure of 200 kPa. A piston, sliding in
the cylinder, compresses the gas isothermally un¬
til the volume is 0.80 m 3 . If the area of the piston
is 240 cm 2 , calculate the force on the piston when
the gas is compressed. (5)
10. Gas at a temperature of 180°C has its volume
reduced by a quarter in an isobaric process.
Determine the final temperature of the gas. (5)
11. Some air at a pressure of 3 bar and at a tempera¬
ture of 60°C occupies a volume of 0.08 m 3 . Cal¬
culate the mass of the air, correct to the nearest
gram, assuming the characteristic gas constant for
air is 287 J/(kg K). (5)
12. A compressed air cylinder has a volume of
1.0 m 3 and contains air at a temperature of 24°C
and a pressure of 1.2 MPa. Air is released from
the cylinder until the pressure falls to 400 kPa and
the temperature is 18°C. Calculate
(a) the mass of air released from the container, and
(b) the volume it would occupy at STP.
Assume the characteristic gas constant for
air to be 287 J/(kg K). (10)
13. A platinum resistance thermometer has a resistance
of 24 Q at 0°C. When measuring the temperature
of an annealing process a resistance value of
68 Q is recorded. To what temperature does this
correspond? Take the temperature coefficient of
resistance of platinum as 0.003 8/°C (5)
14. State which device would be most suitable to
measure the following:
(a) materials in a high-temperature furnace in the
range 1800°C to 3000°C
(b) the air in a factory in the range 0°C to 35°C
(c) an inexpensive method for food processing
applications in the range -20°C to -80°C
(d) boiler flue gas in the range 15°C to 250°C.
(4)
For lecturers/instructors/teachers, fully worked solutions to each of the problems in Revision Test 9,
together with a full marking scheme, are available at the website:
www.routledge.com/cw/bird
A List of formulae for mechanical engineering principles
Formula
Formula symbols
Units
C frr .,„ _ a PP lied force
F
("7 —
Pa
cross-sectional area
U
A
stmin _ change in length
original length
X
e = —
L
stress
Young’s modulus of elasticity =
strain
<7
E= —
£
Pa
force
Stiffness =
extension
k=L
8
N/m
shear stress
Modulus of rigidity = . .
shear strain
G= -
7
Pa
Thermal strain = coefficient of linear expansion
e = aT
x temperature rise
Thermal stress in compound bar
a _ ( a \ ~ E1 ^2 ^2 T
1 (A, E ]+ A 2 E 2 )
Pa
maximum load
Ultimate tensile strength — . . , . ,
° original cross-sectional area
Pa
Moment = force x perpendicular distance
M = Fd
Nm
stress bending moment
distance from neutral axis second moment of area
o _ M _E
J~T~R
N/m 3
Young’s modulus
radius of curvature
Torque = force x perpendicular distance
T = Fd
Nm
Power = torque x angular velocity
P = Tco = 2nnT
W
Horsepower
1 hp = 745.7 W
Torque = moment of inertia x angular acceleration
ll
Nm
shear stress torque
CD
N/m 3
radius polar second moment of area
r J L
(rigidity)(angle of twist)
length
distance travelled s
Average velocity =-:-:- v = - m/s
time taken t
Part Four
Part Four
324 Mechanical Engineering Principles
Formula
Formula symbols
Units
. . . change in velocity
Acceleration = - : -
time taken
a
v — u
t
m/s 2
Linear velocity
V
= cor
m/s
Angular velocity
CO
e _
= — = ZJtn
t
rad/s
Linear acceleration
a
= ra
m/s 2
v ? = Vj + at
m/s
Relationships between initial velocity u , final velocity v,
displacement s, time t and constant acceleration a
<
1 ?
s = ut H— at A
2
v 2 = u 2 + las
m
(m/s) 2
Relationships between initial angular velocity co { , final
angular velocity co 2 , angle 0 , time t and angular accel¬
eration a
<
co 2 = co { + at
1 0
6 = (D\t + —at 1
1 2
co 2 = cof + 2 aO
rad/s
rad
(rad/s) 2
Momentum = mass x velocity
kg m/s
Impulse = applied force x time = change in momentum
kg m/s
Force = mass x acceleration
F
= ma
N
Weight = mass x gravitational field
W = mg
N
Centripetal acceleration
a
v 2
r
m/s 2
Centripetal force
F
2
_ mv
r
N
mass
m
kg/m 3
Denslt y = volume
p
~ V
Work done = force x distance moved
W = Fs
J
„ ~~ . useful output energy
Efficiency = -
input energy
energy used (or work done)
Power = . = force x velocity
time taken
P
E
= — =Fv
t
W
Potential energy = weight x change in height
E p = mgh
J
1 9
kinetic energy = — x mass x (speed)"
E
1 2
t = 2 mv
J
A List of formulae for mechanical engineering principles 325
Formula
Formula symbols
Units
kinetic energy of rotation
= — x moment of inertia x (angular velocity) 2
E k = 2 /+
J
Frictional force = coefficient of friction x normal force
Angle of repose, 0 , on an inclined plane
Efficiency of screw jack
F = juN
tan 0= fi
tan#
rj
SHM periodic time T=2n,
T=2k,
'displacement
acceleration
T =
tan(A + 0 )
2 nj-
a
mass
stiffness
T=
2 *'t
Force ratio =
load
effort
, . , distance moved by effort
Movement ratio = — : ---
distance moved by load
force ratio
Efficiency =-—
movement ratio
Kelvin temperature = degrees Celsius + 273
Quantity of heat energy = mass x specific heat capacity q = mc (t 2 -1 { )
x change in temperature
New length = original length + expansion
New surface area = original surface
area + increase in area
New volume = original volume + increase in volume
force
Pressure =
area
= density x gravitational acceleration x height
Absolute pressure = gauge pressure + atmospheric
pressure
L 2 [1 + Ct{t 2 ^)]
A 2 =A 1 [l+P(t 2 -t 1 )]
V 2 =V l [l+y(t 2 -t l )]
F
p= 7
p=pgh
1 bar = 10 5 Pa
N
simple pendulum
T = 2n'i-
s
compound pendulum
T- 2J<*« + * 2 >
V Sh
s
m
nr
nr
Pa
Part Four
Part Four
326 Mechanical Engineering Principles
1 Formula
Formula symbols
Units
Metacentric height, GM
Px
GM- ‘ cot 0
m
W
Bernoulli’s equation
P { V 2
‘ + ~ + gZl
p 2
P~ V?
- “ + „ +g(z 2 + h f)
p z
Coefficient of discharge
C d = c v x C c
Characteristic gas equation T/ T/
Pl V l - P 2 V 2 - h
Tx T 2
pV= mRT
Circular segment
r2
In Figure FI, shaded area = -—(a ~ sin a)
Figure FI
A List of formulae for mechanical engineering principles 327
Summary of standard results of the second moments of areas of regular sections
Shape
Position of axis
Second moment of area, /
Radius of gyration, k 111
Rectangle
length d
(1) Coinciding with b
bd 3
d
breadth b
3
V3
(2) Coinciding with d
db 3
b
3
(3) Through centroid, parallel to b
bd 3
d
12
^Jl2
(4) Through centroid, parallel to d
db 3
b
12
Vl2
Triangle
Perpendicular
height h
(1) Coinciding with b
bh 3
12
h
S
base b
(2) Through centroid, parallel to base
bh 3
h
36
VI8
(3) Through vertex, parallel to base
bh?
h
4
Circle
radius r
(1) Through centre perpendicular to
7rr 4 nd 4
- or-
r
diameter d
plane (i.e. polar axis)
2 32
(2) Coinciding with diameter
7ir 4 nd 4
- or -
r
4 64
2
(3) About a tangent
Qr 5 nd 4
— r
4 64
2
Semicircle
radius r Coinciding with diameter 1 —
8 2
For a copy of the List of formulae, go to the website:
www.routledge.com/cw/bird
Part Four
Part Four
Metric to Imperial conversions and vice versa
Metric
US or Imperial
1 millimetre, mm
0.03937 inch
1 centimetre, cm = 10 mm
0.3937 inch
1 metre, m = 100 cm
1.0936 yard
1 kilometre, km = 1000 m
0.6214 mile
Metric
US or Imperial
1 cm 2 = 100 mm 2
0.1550 in 2
1 m 2 = 10,000 cm 2
1.1960 yd 2
1 hectare, ha = 10,000 m 2
2.4711 acres
1 km 2 = 100 ha
0.3861 mile 2
Metric
US or Imperial
1 cm 3
0.0610 in 3
1 dm 3 = 1000 cm 3
0.0353 ft 3
1 m 3 = 1000 dm 3
1.3080 yd 3
1 litre = 1 dm 3 = 1000 cm 3
2.113 fluid pt
= 1.7598 pt
Metric
US or Imperial
1 g = 1000 mg
0.0353 oz
1 kg= 1000g
2.2046 lb
1 tonne, t, = 1000 kg
1.1023 short ton
1 tonne, t, = 1000 kg
0.9842 long ton
US or Imperial
Metric
1 inch, in
2.54 cm
1 foot, ft = 12 in
0.3048 m
1 yard, yd = 3 ft
0.9144 m
1 mile = 1760 yd
1.6093 km
1 nautical mile = 2025.4 yd
1.852 km
US or Imperial
Metric
1 in 2
6.4516 cm 2
1 f t 2 = 144 in 2
0.0929 m 2
1 yd 2 = 9 ft 2
0.8361 m 2
1 acre = 4840 yd 2
4046.9 m 2
1 mile 2 = 640 acre
2.59 km 2
US or Imperial
Metric
1 in 3
16.387 cm 3
1 ft 3
0.02832 m 3
1 US floz= 1.0408 UK fl oz
0.0296 litre
1 US pint (16 floz) = 0.8327 UK pt 0.4732 litre
1 US gal (231 in 3 ) = 0.8327 UK gal 3.7854 litre
US or Imperial
Metric
1 oz = 437.5 grain
28.35 g
1 lb = 16 oz
0.4536 kg
1 stone = 14 lb
6.3503 kg
1 hundredweight, cwt = 112 lb
50.802 kg
1 short ton
0.9072 tonne
1 long ton
1.0160 tonne
Greek alphabet
Letter
Upper Case
Lower Case
Alpha
A
a
Beta
B
P
Gamma
r
y
Delta
A
5
Epsilon
E
s
Zeta
Z
c
Eta
H
rj
Theta
0
0
Iota
I
L
Kappa
K
K
Lambda
A
A
Mu
M
Nu
N
V
Xi
w
L_J
S
Omicron
o
o
Pi
n
Jt
Rho
p
p
Sigma
X
o
Tau
T
r
Upsilon
Y
V
Phi
O
<t>
Chi
X
X
Psi
w
ip
Omega
Q
0)
For a copy of Greek Alphabet, go to the website:
www.routledge.com/cw/bird
Part Four
Part Four
GJossary of terms
Acceleration: The amount by which the velocity of an object
increases in a certain time.
Acceleration of free fall: The acceleration experienced by
bodies falling freely in the Earth’s gravitational field.
It varies from place to place around the globe, but is
assigned a standard value of 9.80665m/s 2 , called ‘g’.
Ignoring air resistance, the acceleration does not vary
with the size or shape of the falling body. The value of
‘g’ on the equator ~ 9.78 m/s 2 is less than its value at the
poles, where g ~ 9.83 m/s 2 .
Anemometer: An instrument for measuring wind speed.
It consists of three cups affixed to an upright length of
metal, which in turn drives a mechanism that adjusts a
dial. The cups are blown round by the wind, and the speed
of the wind can be read from the dial.
Angular acceleration: The rate of change of angular
velocity.
Angular momentum: The product of the moment of inertia
I and the angular velocity co of an object.
Angular velocity: The rate of change of an object’s angular
position relative to a fixed point.
Archimedes’ principle: A body immersed in a fluid is
pushed up by a force equal to the weight of the displaced
fluid.
Atmospheric pressure: The downward force exerted by the
atmosphere because of its weight, (gravitational attrac¬
tion to the Earth), measured by barometers, and usually
expressed in units of millibars. Standard atmospheric
pressure at sea level is 1013.25 mb.
Bar: Unit of pressure - the pressure created by a column
of mercury 75.006 cm high at 0°C, or about 33.45 feet of
water at 4°C. It is equal to 10 5 pascal. Standard atmospheric
pressure (at sea level) is 1.01325 bar, or 1013.25 mb.
Barometer: An instrument for measuring atmospheric pres¬
sure. There are two main types - the mercury barometer,
and the aneroid barometer.
Bernoulli’s law: For a steadily flowing fluid (liquid or gas),
the sum of the pressure, kinetic energy per unit volume
and potential energy per unit volume is constant at any
point in the fluid. Using this relationship, it is possible to
measure the velocity of a fluid by measuring its pressure
at two points, as with a manometer or Pitot tube.
Boyle’s law: The volume of a gas at constant temperature is
inversely proportional to the pressure. This means that as
pressure increases, the volume of a gas decreases.
Buoyancy: The upward pressure exerted on an object by the
fluid in which it is immersed. The object is subjected to
pressure from all sides, but the pressure on its lower part
is greater because of the increasing depth of the fluid. The
result of all these pressures is a force acting upwards that
is equal to the weight of the fluid displaced.
Calorie: A unit of heat. A calorie is the amount of heat
required to raise 1 g of water by 1°C between the tem¬
peratures of 14.5°C and 15.5°C. The SI system uses the
joule (1 calorie = 4.184 joules) instead of the calorie.
1000 gram calories = 3.968 Btu (British thermal unit).
1 J = 1 N m.
Celsius: The temperature scale based on the freezing point
of water (0°C) and the boiling point of water (100°C). The
interval between these points is divided into 100 degrees.
The scale was devised by Anders Celsius.
Centre of gravity: Point at which the weight of a body can
be considered to be concentrated and around which its
weight is evenly balanced. In a uniform gravitational
field, the centre of gravity is the same as the centre of
mass.
Centripetal force: In circular or curved motion, the force
acting on an object that keeps it moving in a circular path.
For example, if an object attached to a rope is swung in
a circular motion above a person’s head, the centripetal
force acting on the object is the tension in the rope. Simi¬
larly, the centripetal force acting on the Earth as it orbits
the sun is gravity. In accordance with Newton’s laws, the
reaction to this can be regarded as a centrifugal force,
equal in magnitude and opposite in direction.
Change of state: The change that takes place when matter
turns from one physical phase (gas, liquid or solid) into
another.
Charles’ law: The volume of a gas at constant pressure is
directly proportional to its absolute temperature.
Coefficient of cubic expansion: The fractional increase in
volume per unit temperature rise.
Coefficient of friction: The number characterising the
force necessary to slide or roll one material along
the surface of another. If an object has a weight N and
the coefficient of friction is p, then the force F necessary
to move it without acceleration along a level surface is
F = pN. The coefficient of static friction determines the
force necessary to initiate movement; the coefficient of
kinetic friction determines the force necessary to main¬
tain movement. Kinetic friction is usually smaller than
static friction.
Coefficient of linear expansion: The fractional increase in
length per unit temperature rise.
Glossary of terms 331
Coefficient of superficial expansion: The fractional increase
in area per unit temperature rise.
Conduction, thermal: The transfer of heat from a hot region
of a body to a cold region.
Conservation of energy, law of: States that energy cannot
be created or destroyed.
Convection: The transfer of heat by flow of currents within
fluids due to kinetic theory.
Couple: Two equal and opposite parallel forces, which do
not act in the same line. The forces produce a turning
effect or torque.
Dalton’s law: The pressure exerted by each gas in a mixture
of gases does not depend on the pressures of the other
gases, provided no chemical reaction occurs. The total
pressure of such a mixture is therefore the sum of the par¬
tial pressures exerted by each gas (as if it were alone in the
same volume as the mixture occupies).
Density: The ratio of mass to volume for a given substance
expressed in SI units as kilograms per cubic metre. The
symbol for density is p (Greek rho).
Ductility: Ability of metals and some other materials to be
stretched without being weakened.
Dynamics: The branch of mechanics that deals with
objects in motion. Its two main branches are kinemat¬
ics, which studies motion without regards to its cause,
and kinetics, which also takes into account forces that
cause motion.
Efficiency: The work a machine does (output) divided by the
amount of work put in (input), usually expressed as a per¬
centage. For simple machines, efficiency can be defined
as the force ratio (mechanical advantage) divided by the
distance ratio (velocity ratio).
Elasticity: Capability of a material to recover its size and
shape after deformation by stress. When an external force
(stress) is applied, the material develops strain (a change
in dimension). If a material passes its elastic limit, it will
not return to its original shape.
Energy: The capacity for doing work; it is measured in joules.
Equilibrium: A stable state in which forces acting on a par¬
ticle or object negate each other, resulting in no net force.
Expansion: A change in the size of an object with change
in temperature. Most substances expand on heating,
although there are exceptions - water expands when it
cools from 4°C to its freezing point at 0°C.
Fahrenheit: The temperature scale based on the freez¬
ing point of water (32°F) and the boiling point of water
(212°F). The interval between these points is divided into
180 equal parts. Although replaced by the Celsius scale,
the Fahrenheit scale is still sometimes used for non-
scientific measurements.
Fluid: Any substance that is able to flow. Of the three com¬
mon states of matter, gas and liquid are considered fluid,
while any solid is not.
Force: A push, a pull or a turn. A force acting on an
object may (i) balance an equal but opposite force
or a combination of forces to maintain the object in
equilibrium (so that it does not move), (2) change the state
of motion of the object (in magnitude or direction), or (3)
change the shape or state of the object. The unit of force
is the newton.
Force ratio: The factor by which a simple machine multi¬
plies an applied force. It is the ratio of the load (output
force) to the effort (input force).
Free fall: The state of motion of an unsupported body in a
gravitational field.
Freezing point: The temperature at which a substance
changes phase (or state) from liquid to solid. The freezing
point for most substances increases as pressure increases.
The reverse process, from solid to liquid, is melting; melt¬
ing point is the same as freezing point.
Friction: The resistance encountered when surfaces in con¬
tact slide or roll against each other, or when a fluid (liq¬
uid or gas) flows along a surface. Friction is directly pro¬
portional to the force pressing the surfaces together and
the surface roughness. Before the movement begins, it is
opposed by static friction up to a maximum ‘limiting fric¬
tion’ and then slipping occurs.
Fulcrum: Point about which a lever pivots.
Gear wheel: Is usually toothed, attached to a rotating shaft.
The teeth of one gear engage those of another to transmit
and modify rotary motion and torque. The smaller mem¬
ber of a pair of gears is called a pinion. If the pinion is
on the driving shaft, speed is reduced and turning force
increased. If the larger gear is on the driving shaft, speed
is increased and turning force reduced. A screw-type driv¬
ing gear, called a worm, gives the driven gear a greatly
reduced speed.
Gravity: The gravitational force of attraction at the surface
of a planet or other celestial body. The Earth’s gravity pro-
duces an acceleration of around 9.8 m / s“for any unsup¬
ported body.
Heat: A form of energy associated with the constant vibra¬
tion of atoms and molecules.
Hooke’s law: Within the limit of proportionality, the exten¬
sion of a material is proportional to the applied force.
Approximately, it is the relationship between stress and
strain in an elastic material when it is stretched. The law
states that the stress (force per unit area) is proportional to
the strain (a change in dimensions). The law, which holds
only approximately and over a limited range, was discov¬
ered in 1676 by Robert Hooke.
Horsepower: Unit indicating the rate at which work is done.
The electrical equivalent of one horsepower is 746 watts.
Hydraulics: The physical science and technology of the
behaviour of fluids in both static and dynamic states. It
deals with practical applications of fluid in motion and
devices for its utilisation and control.
Hydrostatics: The branch of mechanics that deals with
liquids at rest. Its practical applications are mainly in
water engineering and in the design of such equipment
as hydraulic presses, rams, lifts and vehicle braking and
control systems.
Part Four
Part Four
332 Mechanical Engineering Principles
Ideal gas laws: The law relating pressure, temperature and
volume of an ideal (perfect) gas pV = mRT, where R is the
gas constant. The law implies that at constant tempera¬
ture T, the product of pressure p and volume V is constant
(Boyle’s law), and at constant pressure, the volume is pro¬
portional to the temperature (Charles’ law).
Imperial system: The units of measurement developed in
the UK. Formerly known as the fps system, which is
an abbreviation for the ‘foot-pound-second system’ of
units.
Inertia: The property possessed by all matter that is a mea¬
sure of the way an object resists changes to its state of
motion.
Joule: The SI unit of energy. One joule is the work done by
a force of one newton acting over a distance of one metre.
The symbol is J, where 1 J = 1 N m.
Kelvin: The SI unit of temperature. The Kelvin temperature
scale has a zero point at absolute zero and degree intervals
(keIvins) the same size as degrees Celsius. The freezing
point of water occurs at 273K (0°C) and the boiling point
at 373 K (100°C).
Kinetic energy: Energy that an object possesses because it
is in motion. It is the energy given to an object to set it
in motion. On impact, it is converted into other forms of
energy such as strain, heat, sound and light.
Latent heat: The heat absorbed or given out by a sub¬
stance as it changes its phase (of matter) at constant
temperature - from a solid to a liquid state or from a
liquid to a gas.
Latent heat of fusion: The heat necessary to transform ice
into water at constant temperature.
Latent heat of vaporisation: The heat necessary to trans¬
form water into steam at constant temperature.
Lever: A simple machine used to multiply the force applied
to an object, usually to raise a heavy load. A lever consists
of a rod and a point (fulcrum) about which the rod pivots.
In a crowbar, for example, the applied force (effort) and
the object to be moved (load) are on opposite sides of the
fulcrum, with the point of application of the effort farther
from it. The lever multiplies the force applied by the ratio
of the two distances.
Machine: A device that modifies or transmits a force in order
to do useful work. In a simple machine, a force (effort)
opposes a larger force (load). The ratio of the load (out¬
put force) to the effort (input force) is the machine’s force
ratio, formerly called mechanical advantage. The ratio
of the distance moved by the load to the distance moved
by the effort is the distance or movement ratio, formerly
known as the velocity ratio. The ratio of the work done
by the machine to that put into it is the efficiency, usually
expressed as a percentage.
Malleability: Property of materials (or other substances)
that can be permanently shaped by hammering or rolling
without breaking. In some cases, it is increased by raising
temperature.
Manometer: A device for measuring pressure.
Mechanical advantage: The factor by which a simple
machine multiplies an applied force. It is the ratio of the
load (output force) to the effort (input force).
Mechanics: The branch of physics concerned with the
behaviour of matter under the influence of forces. It may
be divided into solid mechanics and fluid mechanics.
Another classification is as statics, the study of matter at
rest, and dynamics, the study of matter in motion.
Metric system: The decimal system of weights and mea¬
sures based on a unit of length called the metre and a unit
of mass called the kilogram. Devised by the French in
1791, the metric system is used internationally (SI units)
and has been adopted for general use by most Western
countries, although the imperial system is still commonly
used in the USA and for certain measurements in Britain.
Moment of inertia: For a rotating object, is the sum of the
products formed by multiplying the elements of mass of
the rotating object by the squares of their distances from
the axis of the rotation. Finding this distribution of mass
is important when determining the force needed to make
the object rotate.
Momentum: The product of the mass and linear velocity
of an object. One of the fundamental laws of physics is
the principle that the total momentum of any system of
objects is conserved (remains constant) at all times, even
during and after collisions.
Motion, laws of: Three laws proposed by Isaac Newton
which form the basis of the classical study of motion and
force. According to the first law, a body resists changes
in its state of motion - a body at rest tends to remain at
rest unless acted upon by an external force, and a body
in motion tends to remain in motion at the same veloc¬
ity unless acted on by an external force. This property is
known as inertia. The second law states that the change in
velocity of a body as a result of a force is directly propor¬
tional to the force and inversely proportional to the mass
of the body. According to the third law, to every action
there is an equal and opposite reaction.
Nautical mile: The unit used to measure distances at sea. It
is defined as the length of one minute of arc of the Earth’s
circumference. The international nautical mile is equal to
1852 m (6076.04 feet), but in the UK it is defined as 6080
feet (1853.18 m). A speed of one nautical mile per hour is
called a knot, a term used both at sea and in flying.
Newton: The SI unit of force with the symbol N. One newton
is the force that gives a mass of one kilogram an accelera¬
tion of one metre per second per second. One kilogram
weighs 9.807 N.
Parallelogram of vectors: A method of calculating the sum
of two vector quantities. The direction and size of the vec¬
tors is determined by trigonometry or scale drawing. The
vectors are represented by two adjacent sides of a paral¬
lelogram and the sum is the diagonal through their point
of intersection.
Pascal: The SI unit of pressure with the symbol Pa. It is equal
to a pressure of 1 newton per square metre.
Glossary of terms 333
Pitot tube: A device for measuring the rate of flow of a fluid,
either liquid or gas. For liquids, the device used is gener¬
ally a manometer, with an open end facing upstream and
the other open end out of the stream. The different pres¬
sures at the two ends cause a liquid to shift position within
the two arms of the tube. For gases, a Pitot tube is gener¬
ally L-shaped, with one end open and pointing towards
the flow of gas and the other end connected to a pressure¬
measuring device. This type of Pitot tube is commonly
used as an air-speed indicator in aircraft.
Potential energy: Is an object’s ability to do work because of
a change in the object’s position or shape.
Power: The rate of doing work or of producing or con¬
suming energy. The unit of power is the watt, W, where
1 W = 1 N m/s.
Pressure: The force on an object’s surface divided by the
area of the surface. The SI unit is the pascal (symbol Pa),
which is 1 newton per square metre. In meteorology, the
millibar, which equals 100 pascals, is commonly used.
1 bar = 10 5 Pa = 14.5 psi.
Principle of moments: A law that states that the moments of
two bodies balanced about a central pivot or fulcrum are
equal (the moment of a body being the product of its mass
and its distance from the pivot).
Pulley: A simple machine used to multiply force or to change
the direction of its application. A simple pulley consists of
a wheel, often with a groove, attached to a fixed structure.
Compound pulleys consist of two or more such wheels,
some movable, that allow a person to raise objects much
heavier than he or she could lift unaided.
Pyrometer: A thermometer for use at extremely high
temperatures, well above the ranges of ordinary thermom¬
eters.
Radian: The angle formed by the intersection of two radii
at the centre of a circle, when the length of the arc cut
off by the radii is equal to one radius in length. Thus, the
radian is a unit of angle equal to 57.296°, and there are 2n
radians in 360°.
Radiation: The transmission of energy by subatomic par¬
ticles of electromagnetic waves.
Refrigeration: The process by which the temperature in a
refrigerator is lowered. In a domestic refrigerator, a refrig¬
erant gas, such as ammonia or chlorofluorocarbon (CFC)
is alternately compressed and expanded. The gas is first
compressed by a pump, causing it to warm up. It is then
cooled in a condenser where it liquefies. It is then passed
into an evaporator where it expands and boils, absorbing
heat from its surroundings and thus cooling the refrigera¬
tor. It is then passed through the pump again to be com¬
pressed.
Relative density: The ratio of the density of one substance
to that of a reference substance (usually water) at the
same temperature and pressure. Formerly called specific
capacity.
Scalar: A quantity that only has magnitude; mass, energy
and speed are examples of scalars.
Screw: A variant of a simple machine, the inclined plane. It is
an inclined plane cut around a cone, usually of metal, in a
helical spiral. When force is exerted radially on the screw,
for example by a screwdriver or the lever of a screw-jack,
the screw advances to an extent determined by its pitch
(the distance between crests of its thread).
Shearing force: The force tending to cause deformation of a
material by slipping along a plane parallel to the imposed
stress.
SI units: The Systeme International d’Unites - the interna¬
tionally agreed system of units, derived from the MKS
system (metre, kilogram and second). The seven basic
units are: the metre (m), kilogram (kg), second (s), ampere
(A), kelvin (K), mole (mol), and candela (cd).
Simple harmonic motion (SHM): A periodic motion such
as that of a pendulum, atomic vibrations, or an oscillat¬
ing electric circuit. A body has simple harmonic motion
when it oscillates along a line, moving an equal distance
on either side of a central point and accelerating towards
that point with a speed proportional to its distance from it.
Specific heat capacity: The heat necessary to raise the tem¬
perature of 1 kg of a substance by 1 K. It is measured in
J/(kg K).
Statics: The study of matter at rest. In statics, the forces
on an object are balanced and the object is said to be in
equilibrium; static equilibrium may be stable, unstable
or neutral.
STP: An abbreviation of standard temperature and pressure.
Strain: Change in dimensions of an object subjected to stress.
Linear strain is the ratio of the change in length of a bar to
its original length. Shearing strain describes the change in
shape of an object whose opposite faces are pushed in dif¬
ferent directions. Hooke’s law for elastic materials states
that strain is approximately proportional to stress up to the
material’s limit of proportionality.
Stress: Force per unit area applied to an object. Tensile stress
stretches an object, compressive stress squeezes it, and
shearing stress deforms it sideways. In a fluid, no shear¬
ing stress is possible because the fluid slips sideways, so
all fluid stresses are pressures.
Temperature: A measure of the hotness or coldness of an
object.
Tensile strength: The resistance that a material offers to
tensile stress. It is defined as the smallest tensile stress
required to break the body.
Thermistor: A type of semiconductor whose resistance
sharply decreases with increasing temperature. At 20°C
the resistance may be of the order of a thousand ohms and
at 100°C it may be only ten ohms. Thermistors are used to
measure temperature and to compensate for temperature
changes in other parts of the circuit.
Thermocouple: A thermometer made from two wires of
different metals joined at one end, with the other two
ends maintained at constant temperature. The junction
between the wires is placed in the substance whose tem¬
perature is to be measured. An e.m.f. is generated which
Part Four
Part Four
334 Mechanical Engineering Principles
can be measured and which is, in turn, a measure of tem¬
perature.
Thermopile: A device used to measure radiant heat, consist¬
ing of several thermocouples connected together in series.
Alternate junctions are blackened for absorbing radiant
heat, the other junctions are shielded from the radia¬
tion. The e.m.f. generated by the temperature difference
between the junctions can be measured. From this, the
temperature of the blackened junctions can be calculated,
and thus the intensity of the radiation measured.
Torque: Turning effect of a force. An example is a turbine
that produces a torque on its rotating shaft to turn a gen¬
erator. The unit of measurement is newton metre (N m).
Torsion: Is the strain in material that is subjected to a twist¬
ing force. In a rod or shaft, such as an engine drive shaft,
the torsion angle of twist is inversely proportional to the
fourth power of the rod diameter multiplied by the shear
modulus (a constant) of the material. Torsion bars are
used in the spring mechanism of some car suspensions.
Triangle of vectors: A triangle, the sides of which repre¬
sent the magnitude and direction of three vectors about
a point that are in the same plane and are in equilibrium.
A triangle of vectors is often used to represent forces or
velocities. If the magnitude and direction of two forces
are known, then two sides of the triangle can be drawn.
Using scale drawing or trigonometry, the magnitude and
direction of the third force can be calculated.
Truss: A structural member made up of straight pieces of
metal or timber formed into a series of triangles lying in
a vertical plane. The triangles resist distortion through
stress, making the truss capable of sustaining great loads
over long spans. The joints of a truss are assumed to be
frictionless, pin-jointed.
Vacuum flask: A container for keeping things (usually
liquids) hot or cold. A vacuum flask is made with double
silvered glass walls separated by a near vacuum. The vac¬
uum prevents heat transfer by conduction and convection,
and the silvering on the glass minimises heat transfer by
radiation.
Vaporisation: The conversion of a liquid or solid into its
vapour, such as water into steam.
Vector: A quantity that has both magnitude (size) and direc¬
tion; velocity, acceleration and force are examples of vec¬
tors.
Velocity: The rate of motion of a body in a certain direction.
Velocity ratio: In a simple machine, is the distance moved by
the point of application of the effort (input force) divided
by the distance moved by the point of application of the
load (output force).
Venturi tube: A mechanism for mixing a fine spray of liquid
with a gas for measuring fluid flow.
Watt: The SI unit of power. A machine consuming one joule
of energy per second has a power output of one watt.
1 W = 1 J/s = 1 N m/s. One horsepower corresponds to
746 watts.
Weight: The force of attraction on a body due to gravity.
A body’s weight is the product of its mass and the gravita¬
tional field strength at that point. Mass remains constant,
but weight depends on the object’s position on the Earth’s
surface, decreasing with increasing altitude.
Work: The energy transferred in moving the point of appli¬
cation of a force. It equals the magnitude of the force mul¬
tiplied by the distance moved in the direction of the force.
Young’s modulus: Ratio of the stress exerted on a body to
the longitudinal strain produced.
Answers to multiple-choice questions
Chapter 1
(Exercise 10, Page 16)
Chapter 8
(Exercise 54, Page 124)
1 . (b) 2. (d)
6 . (b) 7. (c)
11 . (b) 12. (d)
16. (c) 17. (d)
3. (a)
8 . (a)
13. (a)
18. (b)
4. (d)
9. (b)
14. (c)
19. (a)
5. (a)
10 . (c)
15. (a)
20 . (c)
1 . (b) 2. (c)
6 . (b)
Chapter 9
3. (c)
4. (a)
5. (c)
142)
(Exercise 60, Page
Chapter 2
(Exercise 20, Page 41)
1 . (c) 2. (b)
3. (d)
4. (a)
5. (b)
6 . (c) 7. (a)
8 . (c)
9. (d)
10 . (a)
1 . (b) 2. (d)
3. (d)
4. (a)
5. (b)
6 . (d) 7. (a)
8 . (a)
9. (d)
10 . (c)
11 . (b) 12. (d)
13. (c)
14. (a)
15. (b)
Chapter 10
16. (c) 17. (c)
18. (a)
19. (c)
20 . (b)
(Exercise 63, Page
150)
1 . (b) 2. (b)
3. (c)
Chapter 3
(Exercise 27, Page 62)
Chapter 11
1 . (c) 2. (c)
3. (a)
4. (b)
5. (c)
(Exercise 69, Page
160)
6 . (c) 7. (b)
8 . (d)
9. (b)
10 . (c)
11 . (f) 12. (h)
13. (d)
14. (b)
15. (a)
1 . (d) 2. (b)
3. (c)
4. (a)
5. (c)
6 . (d) 7. (a)
8 . (b)
9. (c)
10 . (d)
Chapter 4
11 . (a) 12. (c)
(Exercise 31, Page 70)
1 . (f) 2. (d)
3- (g)
4. (b)
Chapter 12
(Exercise 72, Page
166)
Chapter 5
1 . (a) 2. (b)
3. (a)
4. (c)
5. (b)
(Exercise 39, Page 84)
6 . (a)
1 . (b) 2. (a)
3. (b)
4. (d)
5. (b)
6 . (c) 7. (b)
8 . (b)
9. (c)
10 . (d)
Chapter 13
11 . (c) 12. (d)
13. (d)
14. (a)
(Exercise 78, Page
178)
Chapter 6
1 . (b) 2. (c)
3. (a)
4. (c)
5. (a)
6 . (d) 7. (c)
8 . (b)
9. (d)
10 . (c)
(Exercise 45, Page 95)
11 . (b) 12. (d)
13. (a)
1 . (a) 2. (c)
3. (a)
4. (d)
5. (a)
6 . (d) 7. (c)
8 . (a)
9. (d)
10 . (c)
Chapter 14
11 . (c)
(Exercise 82, Page
186)
Chapter 7
(Exercise 50, Page 110)
1. (b) 2. (a) 3. (c) 4. (c) 5. (b)
6 . (a)
1 . (d)
6 . (c)
11 . (b)
2 - (b)
7. (a)
12 . (e)
3. (f)
8 - (g)
4. (c)
9. (f)
5. (a)
10 . (f)
Part Four
Part Four
Chapter 15
6 . (b) 7. (b)
8 . (a)
9. (c)
10 . (b)
(Exercise 87, Page 195)
11 . (d) 12. (c)
13. (d)
1 . (c) 2. (b)
3. (a)
4. (d)
5. (a)
6 . (b) 7. (b)
8 . (a)
9. (a)
10 . (d)
Chapter 22
11 . (d) 12. (c)
13. (b)
(Exercise 125, Page 269)
1 . (b) 2. (c)
3. (a)
4. (d)
5. (b)
Chapter 16
6 . (c) 7. (c)
8 . (a)
9. (c)
10 . (b)
(Exercise 94, Page 209)
1 . (b) 2. (c)
3. (c)
4. (a)
5. (d)
Chapter 23
6 . (c) 7. (a)
8 . (d)
9. (c)
10 . (b)
(Exercise 133, Page 288)
11 . (b) 12. (a)
16. (c)
13. (d)
14. (a)
15. (d)
1 . (b) 2. (d)
3. (a)
4. (a)
5. (c)
6 . (d) 7. (b)
8 . (c)
9. (c)
10 . (d)
11 . (d) 12. (d)
13. (c)
14. (b)
15. (c)
Chapter 17
16. (a) 17. (b)
18. (f)
19. (a)
20 . (b)
(Exercise 99, Page 221)
21 . (c) 22. (c)
23. (a)
24. (c)
1 . (c) 2. (c)
3. (f)
4. (e)
5. (i)
6 . (c) 7. (h)
8 . (b)
9. (d)
10 . (a)
Chapter 24
11 . (b)
(Exercise 138, Page 300)
1 . (c) 2. (c)
3. (d)
4. (a)
5. (c)
Chapter 18
6 . (j) 7. (i)
8 . (d)
9. (g)
10 . (f)
(Exercise 104, Page 230)
11 . (a) 12. (h)
13. (e)
14. (k)
15. (c)
16. (c)
1 . (b) 2. (d)
3. (a)
4. (b)
5. (b)
Chapter 19
Chapter 25
(Exercise 146, Page 310)
(Exercise 108, Page 237)
1 . (b) 2. (c)
3. (b)
4. (a)
1 . (a) 2. (d)
3. (b)
4. (b)
5. (c)
6 . (d) 7. (b)
8 . (c)
9. (c)
10 . (b)
Chapter 20
Chapter 26
(Exercise 115, Page 247)
(Exercise 150, Page 320)
1 . (b) 2. (f)
3. (c)
4. (d)
5. (b)
1 . (c) 2. (b)
3. (d)
4. (b)
5. (i)
6 . (a) 7. (b)
8 . (d)
9. (c)
10 . (d)
6 . (a) 7. (e)
8 . (d)
9. (e or
f) 10. (k)
11 . (d) 12. (b)
11 . (b) 12. (g)
Chapter 21
(Exercise 121, Page 261)
4. (c) 5. (b)
1 . (d) 2. (b)
3. (a)
Index
Absolute pressure, 275, 280
Acceleration, centripetal, 192
linear and angular, 173
Aneroid barometer, 279
Angle, measurement of, 4
of repose, 215
of twist, 161
Angular acceleration, 173
motion, 171
velocity, 171
Annulus, 135
Applications of friction, 214
thermal expansion, 264
Archimedes’ principle, 276
Atmospheric pressure, 275, 280
Bar, 278
Barometer, 278
Beckman thermometer, 313
Belt, 157
Bending moment, 112, 122
diagram, 113, 122
Bending of beams, 145
Bernoulli’s equation, 297
Bimetallic thermometer, 319
Bourdon pressure gauge, 281
Bow’s notation, 100
Boyle’s law, 301
Brackets, 8
Brittleness, 57
Built-up sections, 138
Buoyancy, 284
Calculator, 9
Calculus, 34
Cartesian axes, 28
Celsius scale, 254
Centre of gravity, 72, 127
Centrifugal clutch, 230
force, 224
Centripetal acceleration, 192
force, 192
Centroidal axis, 148
Centroids, 127, 128
Change of state, 256
temperature, 263
Channel bar, 138
Characteristic equation, 306
gas constant, 306
Charles’ law, 303
Clinical thermometer, 313
Clutch, 230
Coefficient of cubic expansion, 267
discharge, 298
friction, 213
linear expansion, 264
superficial expansion, 266
Compatibility, 58
Compound bars, 58
gear train, 244
pendulum, 236
Compression, 48
Compressive force, 48
Concurrent forces, 72
Conduction, 259
Conical pendulum, 226
Contraction, 263
Convection, 259
Co-ordinates, 28
Coplanar forces, 72
in equilibrium, 78
resultant of, 73
Cosine, 5
rule, 7, 76
Couple, 102, 151
Couples, 92
Cross product, 40
Cubic expansion, coefficient of, 267
Cup anemometers, 294
D’Alembert’s principle, 224
Dalton’s law of partial pressure, 305
Dead weight tester, 282
Definite integrals, 38
Deflecting vane flowmeter, 293
Degrees, 4
Denominator, 8
Derived units, 22
Determinants, 39
Differential calculus, 34
pressure flowmeters, 290
Differentiation, 34
of common functions, 34
Dot product, 40
Ductility, 57
Dynamic friction, 212
coefficient of, 214, 215
Effects of forces on materials, 47
Efficiency, 157, 201
of a screw jack, 219
of a simple machine, 239
Effort, 239
Elastic collisions, 206
limit, 52, 65
Elasticity, 52
Electromagnetic flowmeter, 295
Energy, 201
kinetic, 205
potential, 205
Engineering notation, 23
Equation of a straight line graph, 31
continuity, 296
Equations of motion, 174, 175
Equilibrium, 58, 72, 78, 87
Expansion, 263
and contraction of water, 264
Extrapolation, 29
First moment of area, 127, 128, 131
Float and tapered tube meter, 294
Flowmeters, 290
electromagnetic, 295
mechanical, 293
Flow nozzle, 292
Flow through an orifice, 298
Fluid flow, 290
pressure, 274
Follower, 243
Force, 48, 188
centrifugal, 224
centripetal, 192
gravitational, 189
ratio, 239
Forces, 72
acting at a point, 71
in structures, 98
graphical method, 100
method of joints, 104
method of sections, 109
Formulae, list of, 323-327
Fortin barometer, 279
Fractions, 8
Part Four
338 Mechanical Engineering Principles
Friction, 212
applications of, 214
coefficient of, 213
on an inclined plane, 215
Fulcrum, 87
Gas laws, 301
thermometers, 319
Gauge pressure, 275, 280
Gay-Lussac’s law, 304
Gear trains, 243, 244
Glossary of terms, 330-334
Gradient of straight line, 30
Graph, 28
Gravitational force, 189
Greek alphabet, 329
Heat, 253
Hogging, 113
Hooke’s law, 53
Horizontal component, 80
Horsepower, 152
Hot-wire anemometer, 296
Hydrostatic pressure, 282
thrust on curved surface, 284
Hydrostatics, 272
/beam, 138
Ideal gas, 305
laws, 301
Idler wheel, 244
i, j, k unit vector, 39
Impact of a jet, 298
Imperial to metric conversions, 24, 328
Improper fraction, 8
Impulse, 183
Impulsive forces, 183
Inclined plane, friction on, 215
manometer, 281
Indices, laws of, 12
Inelastic collisions, 206
Inertia, 189
moment of, 194
Insulation, use of conserving fuel, 260
Integral calculus, 34
Integral, definite, 38
Integration, 34, 36
Interpolation, 29
Isobaric process, 303
Isothermal change, 302
Joule, 198
Kelvin, 254
Kinetic energy, 154, 205
of rotation, 208
Lamina, 72, 127
Latent heat, 257
of fusion, 257
of vaporisation, 257
Law of machine, 240
Laws of indices, 12
Levers, 245
Limiting angle of repose, 215
coefficient of friction, 214, 215
efficiency, 240
force ratio, 240
Limit of proportionality, 52, 65
Linear and angular acceleration, 173
Linear expansion, 264
momentum, 180
motion, 171
velocity, 171
Linear expansion, coefficient of,
264
Liquid-in-glass thermometer, 255,
312
Load, 239
Machines, 239
Malleability, 57
Manometer, 280
Mariana Trench, 274
Mathematics revision, 3-42
Maximum/minimum thermometers,
313
McLeod gauge, 282
Measurement of angles, 4
pressure, 278
temperature, 254, 312
Mechanical advantage, 239
flowmeters, 293
Mechanisms, 99
Mercury-in-steel thermometer, 319
Mercury thermometers, 313
Metacentric height, 284
Method of joints, 104
sections, 109
Metric to imperial conversions, 24,
328
Mid-ordinate rule, 198
Mixed numbers, 9
Modulus of elasticity, 56
rigidity, 57
Moment, 40, 86, 102
of a force, 86
of inertia, 154, 194
of resistance, 147
Moments, principle of, 87
Momentum, 180
Motion down a plane, 216
equations of, 174
in a circle, 223
in a vertical circle, 228
on a curved banked track, 225
up a plane, 216
Movement ratio, 239, 243
Neck, 65
Neutral axis, 146, 147
layer, 146
Newton, 22, 48, 188, 189
Newton metre, 87, 151
Newton’s laws of motion, 180, 183,
189
Normal force, 213
Numerator, 8
Optical pyrometer, 318
Orifice plate, 291
Origin, 28
Parallel axis theorem, 132, 194
Parallelogram of forces, 75
Partial pressure, 305
Pascal, 49, 272, 278
Pendulum, compound, 236
conical, 226
simple, 235
Percentages, 10
Permanent elongation, 65
Perpendicular axis theorem, 132
Pirani gauge, 282
Pitometer, 293
Pitot-static tube, 293
Pivot, 87
Plasticity, 52
Platinum coil, 316
Point loading, 89
Polar second moment of area, 161
Polygon of forces, 77
Potential energy, 205
Power, 152,202
transmission, 157
Practical applications of thermal
expansion, 264
straight line graphs, 32
Prefixes, engineering, 22
Pressure, 272
absolute, 275, 280
atmospheric, 275, 280
fluid, 274
gauge, 275, 280
gauges, 280
hydrostatic, 282
law, 304
measurement of, 278
partial, 305
Index 339
Principle of conservation of energy,
180, 201,206
mass, 296
moments, 87
momentum, 180
Proof stress, 69
Proper fraction, 8
Pulleys, 241
Pyrometers, 255, 317
Pythagoras’ theorem, 6
Radian, 4, 171
Radiation, 259
Radius of curvature, 146
gyration, 131
Reaction, 190
Reactions, 90
Rectangular axes, 28
Refrigerator, 259
Relative velocity, 176
Resistance thermometers, 255, 315
Resolution of forces, 76, 80
Resultant, 73
Resultant of coplanar forces, 73, 76
by calculation, 76
by vector addition, 73, 76
Rigidity modulus, 161
Rotameters, 295
Rotary vane positive displacement
meters, 294
Rotation of a rigid body, 193
RSJ, 138
Sagging, 113
Scalar product, 40
quantity, 71, 176
Screw jack, 219, 243
efficiency of, 219
Second moments of area, 131, 146
for built-up sections, 138
table of, 133
Sensible heat, 257
Standard derivatives, 34
integrals, 36
Shear, 48
force, 48
modulus, 161
stress, 161
Shearing force, 113, 122
diagram, 113, 122
Simple harmonic motion (SHM),
232
machines, 239
pendulum, 235
refrigerator, 259
Simply supported beam, 87
having point loads, 89
practical applications, 90
with couples, 93
Simultaneous equations, 14
Sine, 5
rule, 7, 76
SI units, 21
Sliding friction, 212, 215
coefficient of, 214
Specific heat capacity, 255
Spring-mass system, 233
Spur gears, 243
Stability of floating bodies, 284
Standard integrals, 36
temperature and pressure (STP),
306
Statically indeterminate trusses,
99
Static friction, 213, 215
coefficient of, 214, 215
Stiction, 213
Stiffness, 53
Straight line graphs, 28
Strain, 50
thermal, 28
Stress, 49, 146
proof, 69
Stmt, 99
Superficial expansion, coefficient of,
266
Tangent, 5
Tee bar, 138
Temperature, 27, 253
indicating paints, 319
measurement of, 312
sensitive crayons, 319
Tensile force, 48
test, 64
Tension, 48
Testing to destmction, 64
Theorem of Pythagoras, 6
Thermal expansion, 263
practical applications, 264
Thermal strain, 57
Thermistors, 317
Thermocouples, 255, 314
Thermodynamic scale, 254
Thermometer, 254
Tie, 99
Torsional vibrations, 237
Torsion of shafts, 161
Torque, 151, 161
Total radiation pyrometer, 317
Triangle calculations, 5
of forces, 74
Truss, 98
Turbine flowmeter, 294
-type meters, 294
Twisting of shafts, 161
Ultimate tensile strength (UTS), 65
Uniformly distributed loads (UDL),
122
Units, SI, 21
Upper yield point, 65
U-tube manometer, 280
Vacuum flask, 260
gauge, 282
Vector addition, 73
analysis, 39
product, 40
quantity, 71, 176
Velocity, linear and angular, 171
Velocity ratio, 239
Velocity, relative, 176
Venturi tube, 292
Vertical-axis intercept, 29
Vertical component, 80
Waist, 65
Water, expansion and contraction, 264
Watt, 203
Wheatstone bridge circuit, 316
Work, 197
done, 40, 152, 197
y-axis intercept, 31
Yield point, 65
stress, 65
Young’s modulus of elasticity, 33,
53, 146
Part Four
1 UNDERSTANDING
* ENGINEERING
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